A Second Course in Mathematics Concepts for Elementary Teachers

A Second Course in Mathematics Concepts forElementary Teachers: Theory, Problems, and

Solutions

Marcel B. FinanArkansas Tech University

c©All Rights ReservedFirst Draft

February 8, 2006

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Contents

25 Integers: Addition and Subtraction 3

26 Integers: Multiplication, Division, and Order 15

27 Rational Numbers 34

28 Real Numbers 46

29 Functions and their Graphs 54

30 Graphical Representations of Data 70

31 Misleading Graphs and Statistics 88

32 Measures of Central Tendency and Dispersion 98

33 Probability: Some Basic Terms 117

34 Probability and Counting Techniques 131

35 Permutations, Combinations and Probability 143

36 Odds, Expected Value, and Conditional Probability 152

37 Basic Geometric Shapes and Figures 162

38 Properties of Geometric Shapes 170

39 Symmetry of Plane Figures 180

40 Regular Polygons 190

41 Three Dimensional Shapes 196

42 Systems of Measurements 207

43 Perimeter and Area 216

44 Surface Area 229

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45 Volume 238

46 Congruence of Triangles 246

47 Similar Triangles 254

48 Solutions to Practice Problems 266

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25 Integers: Addition and Subtraction

Whole numbers and their operations were developed as a direct result ofpeople’s need to count. But nowadays many quantitative needs aside fromcounting require numbers other than whole numbers.In this and the next section we study the set Z of integers which consists of:• the set W of whole numbers, i.e. W = {0, 1, 2, 3, · · ·},• and the negative integers denoted by N = {−1,−2,−3, · · ·}.

That is,Z = {· · · ,−3,−2,−1, 0, 1, 2, 3, · · ·}.

In today’s society these numbers are used to record debits and credits, profitsand losses, changes in the stock markets, degrees above and below zero inmeasuring temperature, the altitude above and below sea level, etc.As usual, we introduce the set of integers by means of pictures and diagramswith steadily increasing levels of abstraction to suggest, in turn, how theseideas might be presented to school children.

Representations of IntegersWe will discuss two different ways for representing integers: set model andnumber line model.• Set Model: In this model, we use signed counters. We represent +1 by⊕ and −1 by . In this case, the number +2 is represented by

⊕ ⊕

and the number −5 is represented by

One plus counter neutralizes one minus counter. Thus, we can represent 0by ⊕ .• Number-Line Model: The number line model is shown in Figure 25.1

Figure 25.1

Example 25.1Represent −4 on a number line and as a set of signed counters.

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Solution.Using signed counters we have

Figure 25.2 shows the position of −4 on a number-line.

Figure 25.2

Looking at Figure 25.1, we see that −2 and 2 have the same distance from0 but on opposite sides. Such numbers are called opposite. In general, theopposite of x is −x. Hence, the opposite of 5 is −5 and the opposite of −5 is−(−5) = 5.

Example 25.2Find the opposite of each of these integers.(a) 4 (b) −2 (c) 0

Solution.(a) The opposite of 4 is −4.(b) The opposite of −2 is −(−2) = 2.(c) The opposite of 0 is 0 since 0 is neither positive nor negative.

Absolute Value of an IntegerAs an important application of the use of the number line model for integersis the concept of absolute value of an integer. We see from Figure 25.1 thatthe number −3 is represented by the point numbered −3 and the integer 3is represented by the point numbered 3. Note that both −3 and 3 are threeunits from 0 on the number line. The absolute value of an integer x isthe distance of the corresponding point on the number line from zero. Weindicate the absolute value of x by |x|. It follows from the above discussionthat | − 3| = |3| = 3.

Example 25.3Determine the absolute values of these integers.(a) −11 (b) 11 (c) 0

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Solution.(a) | − 11| = 11 (b) |11| = 11 (c) |0| = 0.

Practice Problems

Problem 25.1Which of the following are integers?(a) −11 (b) 0 (c) 3

4(d) −9

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Problem 25.2Let N = {−1,−2,−3, · · ·}. Find(a) N ∪W(b) N ∩N, where N is the set of natural numbers or positive integers and Wis the set of whole numbers.

Problem 25.3What number is 5 units to the left of −95?

Problem 25.4A and B are 9 units apart on the number line. A is twice as far from 0 asB. What are A and B?

Problem 25.5Represent −5 using signed counters and number line.

Problem 25.6In terms of distance, explain why | − 4| = 4.

Problem 25.7Find the additive inverse (i.e. opposite) of each of the following integers.(a) 2 (b) 0 (c) −5 (d) m (e) −m

Problem 25.8Evaluate each of the following.(a) | − 5| (b) |10| (c) −| − 4| (d) −|5|

Problem 25.9Find all possible integers x such that |x| = 2.

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Problem 25.10Let W be the set of whole numbers, Z+ the set of positive integers (i.e.Z+ = N), Z− the set of negative integers, and Z the set of all integers. Findeach of the following.(a) W ∪ Z (b) W ∩ Z (c) Z+ ∪ Z− (d) Z+ ∩ Z− (e) W − Z+

In the remainder of this section we discuss integer addition and subtraction.We will use the devices of signed counters and number lines to illustrate howthese operations should be performed in the set of integers.

Addition of IntegersA football player loses 3 yards on one play and 2 yards on the next, what’sthe player’s overall yardage on the two plays? The answer is −3+(−2) whichinvolves addition of integers. We compute this sum using the two models dis-cussed above.• Using Signed CountersSince −3 is represented by 3 negative counters and −2 is represented by 2negative counters then by combining these counters we obtain 5 negativecounters which represent the integer −5. So −3 + (−2) = −5.• Using Number LineMove three units to the left of 0 and then 2 units to the left from −3 asshown in Figure 25.3.

Figure 25.3

Example 25.4Find the following sums using (i) signed counters, and (ii) number line.(a) (−7) + 4 (b) 5 + (−2) (c) 2 + (−2)

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Solution.

The results of the above examples are entirely typical and we state themas a theorem.

Theorem 25.1Let a and b be integers. Then the following are always true:(i) (−a) + (−b) = −(a + b)(ii) If a > b then a + (−b) = a− b(iii) If a < b then a + (−b) = −(b− a)(iv) a + (−a) = 0.

Example 25.5Compute the following sums.(a) 7 + 11 (b) (−6) + (−5) (c) 7 + (−3) (d) 4 + (−9) (e) 6 + (−6)

Solution.(a) 7 + 11 = 18(b) (−6) + (−5) = −(6 + 5) = −11(c) 7 + (−3) = 7− 3 = 4(d) 4 + (−9) = −(9− 4) = −5(e) 6 + (−6) = 0

Integer addition has all the properties of whole number addition. Theseproperties are summarized in the following theorem.

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Theorem 25.2Let a, b, and c be integers.Closure a + b is a unique integer.Commutativity a + b = b + a.Associativity a + (b + c) = (a + b) + c.Identity Element 0 is the unique integer such that a + 0 = 0 + a = a.

A useful consequence of Theorem 25.1 (iv) is the additive cancellation.

Theorem 25.3For any integers a, b, c, if a + c = b + c then a = b.

Proof.

a = a + 0 identity element= a + (c + (−c)) additive inverse= (a + c) + (−c) associativity= (b + c) + c given= b + (c + (−c)) associativity= b + 0 additive inverse= b identity element

Example 25.6Use the additive cancellation to show that −(−a) = a.

Solution.Since a + (−a) = (−(−a)) + (−a) = 0 then by the additive cancellationproperty we have a = −(−a)

Practice Problems

Problem 25.11What is the opposite or additive inverse of each of the following? (a and bare integers)(a) a + b (b) a− b

Problem 25.12What integer addition problem is shown on the number line?

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Problem 25.13(a) Explain how to compute −7 + 2 with a number line.(b) Explain how to compute −7 + 2 with signed counters.

Problem 25.14In today’s mail, you will receive a check for $86, a bill for $30, and anotherbill for $20. Write an integer addition equation that gives the overall gain orloss.

Problem 25.15Compute the following without a calculator.(a) −54 + 25 (b) (−8) + (−17) (c) 400 + (−35)

Problem 25.16Show two ways to represent the integer 3 using signed counters.

Problem 25.17Illustrate each of the following addition using the signed counters.(a) 5 + (−3) (b) −2 + 3 (c) −3 + 2 (d) (−3) + (−2)

Problem 25.18Demonstrate each of the additions in the previous problem using number linemodel.

Problem 25.19Write an addition problem that corresponds to each of the following sentencesand then answer the question.(a) The temperature was −10◦C and then it rose by 8◦C. What is the newtemperature?(b) A certain stock dropped 17 points and the following day gained 10 points.What was the net change in the stock’s worth?(c) A visitor to a casino lost $200, won $100, and then lost $50. What is thechange in the gambler’s net worth?

Problem 25.20Compute each of the following: (a) |(−9) + (−5)| (b) |7 + (−5)| (c)|(−7) + 6|

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Problem 25.21If a is an element of {−3,−2,−1, 0, 1, 2} and b is an element of {−5,−4,−3,−2,−1, 0, 1},find the smallest and largest values for the following expressions.(a) a + b (b) |a + b|

Subtraction of IntegersLike integer addition, integer subtraction can be illustrated with signed coun-ters and number line.

• Signed CountersTake Away ApproachAs with the subtraction of whole numbers, one approach to integer subtrac-tion is the notion of ”take away.” Figure 25.4 illustrates various examples ofsubtraction of integers.

Figure 25.4

Another important fact about subtraction of integers is illustrated in Figure25.5.

Figure 25.5

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Note that 5− 3 = 5 + (−3). In general, the following is true.

Adding the Opposite ApproachFor all integers a and b, a− b = a + (−b).

Adding the opposite is perhaps the most efficient method for subtractingintegers because it replaces any subtraction problem with an equivalent ad-dition problem.

Example 25.7Find the following differences by adding the opposite.(a) −8− 3 (b) 4− (−5)

Solution.(a) −8− 3 = (−8) + (−3) = −(8 + 3) = −11.(b) 4− (−5) = 4 + [−(−5)] = 4 + 5 = 9.

Missing-Addend ApproachSigned counters model can be also used to illustrate the ”missing-addend”approach for the subtraction of integers. Figure 25.6 illustrates both equa-tions 3− (−2) = 5 and 3 = 5 + (−2).

Figure 25.6

Since all subtraction diagrams can be interpreted in this way, we can intro-duce the ”missing-addend” approach for subtraction:If a, b, and c are any integers, then a− b = c if and only if a = c + b.

• Number Line ModelThe number line model used for integer addition can also be used to modelinteger subtraction. In this model, the operation of subtraction correspondsof starting from the tail of the first number facing the negative direction.

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Thus, from the tail of the first number we move forward if we are subtract-ing a positive integer and backward if we are subtracting a negative integer.Figure 25.7 illustrates some examples.

Figure 25.7

In summary there are three equivalent ways to view subtraction of integers.

1. Take away2. Adding the opposite3. Missing addend

Example 25.8Find 4− (−2) using all three methods of subtraction.

Solution.Take away: See Figure 25.8

Figure 25.8

Adding the opposite: 4− (−2) = 4 + [−(−2)] = 4 + 2 = 6.Missing-Addend: 4− (−2) = c if and only if 4 = c + (−2). But 4 = 6 + (−2)so that c = 6.

Practice Problems

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Problem 25.22Explain how to compute 5− (−2) using a number line.

Problem 25.23Tell what subtraction problem each picture illustrates.

Problem 25.24Explain how to compute the following with signed counters.(a) (−6)− (−2) (b) 2− 6 (c) −2− 3 (d) 2− (−4)

Problem 25.25(a) On a number line, subtracting 3 is the same as moving units tothe .(b) On a number line, adding −3 is the same as moving units to the

.

Problem 25.26An elevetor is at an altitude of −10 feet. The elevator goes down 30 ft.(a) Write an integer equation for this situation.(b) What is the new altitude?

Problem 25.27Compute each of the following using a− b = a + (−b).(a) 3− (−2) (b) −3− 2 (c) −3− (−2)

Problem 25.28Use number-line model to find the following.(a) −4− (−1) (b) −2− 1.

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Problem 25.29Compute each of the following.(a) |5− 11| (b) |(−4)− (−10)| (c) |8− (−3)| (d) |(−8)− 2|

Problem 25.30Find x.(a) x + 21 = 16 (b) (−5) + x = 7 (c) x− 6 = −5 (d) x− (−8) = 17.

Problem 25.31Which of the following properties hold for integer subtraction. If a propertydoes not hold, disprove it by a counterexample.(a) Closure (b) Commutative (c) Associative (d) Identity

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26 Integers: Multiplication, Division, and Or-

der

Integer multiplication and division are extensions of whole number multipli-cation and division. In multiplying and dividing integers, the one new issue iswhether the result is positive or negative. This section shows how to explainthe sign of an integer product or quotient using number line, patterns, andsigned counters.

Multiplication of Integers Using PatternsConsider the following pattern of equalities which are derived from using therepeated addition .

4× 3 = 124× 2 = 84× 1 = 44× 0 = 0

Note that the second factors in the successive products decrease by 1 eachtime and that the successive results decrease by 4. If this pattern continueswe obtain the following sequence of equalities.

4× 3 = 124× 2 = 84× 1 = 44× 0 = 0

4× (−1) = −44× (−2) = −84× (−3) = −12

Since these results agree with what would be obtained by repeated addition,the pattern of the product is an appropriate guide. In general, it suggeststhat

m · (−n) = −(m · n)

where m and n are positive integers. This says that the product of a positiveinteger times a negative integer is always a negative integer.

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Using this result we can write the following pattern of equalities

3× (−3) = −92× (−3) = −61× (−3) = −30× (−3) = 0

So we notice that when the first factors decrease by 1 the successive resultsincrease by 3. Continuing the pattern we find

3× (−3) = −92× (−3) = −61× (−3) = −30× (−3) = 0

(−1)× (−3) = 3(−2)× (−3) = 6

In general, this suggests that

(−m) · (−n) = m · n,

where m and n are positive integers. Hence the product of two negativeintegers is always positive.What about the product of a negative integer times a positive integer suchas (−3)× 2? Using the results just derived above we see that

(−3)× (−3) = 9(−3)× (−2) = 6(−3)× (−1) = 3

(−3)× 0 = 0

Note that the second factors in the successive products increase by 1 eachtime and the successive results decrease by 3. Hence, continuing the patternwe obtain

(−3)× (−3) = 9(−3)× (−2) = 6(−3)× (−1) = 3

(−3)× 0 = 0(−3)× 1 = −3(−3)× 2 = −6

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This suggests the general rule

(−n) ·m = −(n ·m),

where m and n are positive integers. That is, a negative number times apositive number is always negative.Summarizing these results into a theorem we have

Theorem 26.1Let m and n be two positive integers. Then the following are true:(1) m(−n) = −mn(2) (−m)n = −mn(3) (−m)(−n) = mn.

Multiplication of Integers Using Signed CountersThe signed counters can be used to illustrate multiplication of integers, al-though an interpretation must be given to the sign. See Figure 26.1(a). Theproduct 3 × 2 illustrates three groups each having 2 positive counters. Aproduct such as 3 × (−2) represents three groups each having 2 negativecounters as shown in Figure 26.1(b). A product like (−3)× 2 is interpretedas taking away 3 groups each having 2 positive counters as shown in Figure26.1(c), and the product (−3) × (−2) is interpreted as taking away threegroups each consisting of two negative counters as shown in Figure 26.1(d).

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Figure 26.1

Multiplication of Integers Using Number LineSince a product has a first factor and a second factor we can identify the firstfactor as the velocity of a moving object who starts from 0 and the secondfactor is time. A positive velocity means that the object is moving east anda negative velocity means that the object is moving west. Time in future isdenoted by a positive integer and time in the past is denoted by a negativeinteger.

Example 26.1Use a number-line model to answer the following questions.(a) If you are now at 0 and travel east at a speed of 50km/hr, where will yoube 3 hours from now?(b) If you are now at 0 and travel east at a speed of 50km/hr, where wereyou 3 hours ago?(c) If you are now at 0 and travel west at a speed of 50km/hr, where will yoube 3 hours from now?(d) If you are now at 0 and travel west at a speed of 50km/hr, where wereyou 3 hours ago?

Solution.Figure 26.2 illustrates the various products using a number-line model.

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Figure 26.2

The set of integers has properties under multiplication analogous to those ofthe set of whole numbers. We summarize these properties in the followingtheorem.

Theorem 26.2Let a, b, and c be any integers. Then the following are true.

Closure a · b is a unique integer.Commutativity a · b = b · aAssociativity a · (b · c) = (a · b) · cIdentity Element a · 1 = 1 · a = aDistributivity a · (b + c) = a · b + a · c.Zero Product a · 0 = 0 · a = 0.

Using the previous theorem one can derive some important results that areuseful in computations.

Theorem 26.3For any integer a we have (−1) · a = −a.

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Proof.First, note that 0 = 0 · a = [1 + (−1)] · a = a + (−1) · a. But a + (−a) = 0 sothat a+(−a) = a+(−1)a. By the additive cancellation property we concludethat −a = (−1) · a

We have seen that (−a)b = −ab and (−a)(−b) = ab where a and b arepositive integer. The following theorem shows that those two equations aretrue for any integer not just positive integers.

Theorem 26.4Let a and b be any integers. Then

(−a)b = −ab

and(−a)(−b) = ab.

Proof.To prove the first equation, we proceed as follows.

(−a)b = [(−1)a]b = (−1)(ab) = −ab.

To prove the second property, we have

(−a)(−b) = (−1)[a(−b)] = (−1)[(−a)b] = (−1)[(−1)(ab)] = (−1)(−1)(ab) = ab

Another important property is the so-called the multiplicative cancella-tion property.

Theorem 26.5Let a, b, and c be integers with c 6= 0. If ac = bc then a = b.

Proof.Suppose c 6= 0. Then either c > 0 or c < 0. If c

1= c > 0 then we know from

Theorem 20.1 that c has a multiplicative inverse 1c, i.e. cc−1 = 1. In this case,

we havea = a · 1

= a(cc−1)= (ac)c−1

= (bc)c−1

= b(cc−1)= b · 1 = b

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If c < 0 then −c > 0 and we can use the previous argument to obtain

a = a · 1= a[(−c)(−c)−1]= [a(−c)](−c)−1

= −(ac)(−c)−1

= −(bc)(−c)−1

= [b(−c)](−c)−1

= b[(−c)(−c)−1]= b · 1 = b

The following result follows from the multiplicative cancellation property.

Theorem 26.6 (Zero Divisor)Let a and b be integers such that ab = 0. Then either a = 0 or b = 0.

Proof.Suppose that b 6= 0. Then ab = 0 · b. By the previous theorem we must havea = 0

Practice Problems

Problem 26.1Use patterns to show that (−1)(−1) = 1.

Problem 26.2Use signed counters to show that (−4)(−2) = 8.

Problem 26.3Use number line to show that (−4)× 2 = −8.

Problem 26.4Change 3× (−2) into a repeated addition and then compute the answer.

Problem 26.5(a) Compute 4× (−3) with repeated addition.(b) Compute 4× (−3) using signed counters.(c) Compute 4× (−3) using number line.

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Problem 26.6Show how (−2)× 4 can be found by extending a pattern in a whole-numbermultiplication.

Problem 26.7Mike lost 3 pounds each week for 4 weeks.(a) What was the total change in his weight?(b) Write an integer equation for this situation.

Problem 26.8Compute the following without a calculator.(a) 3× (−8)(b) (−5)× (−8)× (−2)× (−3)

Problem 26.9Extend the following pattern by writing the next three equations.

6× 3 = 186× 2 = 126× 1 = 66× 0 = 0

Problem 26.10Find the following products.(a) 6(−5) (b) (−2)(−16) (c) −(−3)(−5) (d) −3(−7− 6).

Problem 26.11Represent the following products using signed counters and give the results.(a) 3× (−2) (b) (−3)× (−4)

Problem 26.12In each of the following steps state the property used in the equations.

a(b− c) = a[b + (−c)]= ab + a(−c)= ab + [−(ac)]= ab− ac

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Problem 26.13Extend the meaning of a whole number exponent

an = a · a · a · · · a︸︷︷︸n factors

where a is any integer and n is a positive integer. Use this definition to findthe following values.(a) (−2)4 (b) −24 (c) (−3)5 (d) −35

Problem 26.14Illustrate the following products on an integer number line.(a) 2× (−5) (b) 3× (−4) (c) 5× (−2)

Problem 26.15Expand each of the following products.(a) −6(x + 2) (b) −5(x− 11) (c) (x− 3)(x + 2)

Problem 26.16Name the property of multiplication of integers that is used to justify eachof the following equations.(a) (−3)(−4) = (−4)(−3)(b) (−5)[(−2)(−7)] = [(−5)(−2)](−7)(c) (−5)(−7) is a unique integer(d) (−8)× 1 = −8(e) 4 · [(−8) + 7] = 4 · (−8) + 4 · 7

Problem 26.17If 3x = 0 what can you conclude about the value of x?

Problem 26.18If a and b are negative integers and c is positive integer, determine whetherthe following are positive or negative.(a) (−a)(−c) (b) (−a)(b) (c) (c− b)(c− a) (d) a(b− c)

Problem 26.19Is the following equation true? a · (b · c) = (a · b) · (a · c).

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Division of IntegersSuppose that a town’s population drops by 400 people in 5 years. What isthe average population change per year? The answer is given by −400 ÷ 5,where the negative sign indicates a decrease in population. Such a problemrequires division of integers which we dicuss next.You recall the missing-factor approach for the division of whole numbers: wewrite a÷b, b 6= 0 to mean a unique whole number c such that a = bc. Divisionon the set of integers is defined analogously:Let a and b be any integers with b 6= 0. Then a÷ b is the unique integer c, ifit exists, such that a = bc.

Example 26.2Find the following quotients, if possible.(a) 12÷ (−4) (b) (−12)÷ 4 (c) (−12)÷ (−4) (d) 7÷ (−2)

Solution.(a) Let c = 12÷ (−4). Then −4c = 12 and consequently, c = −3.(b) If c = (−12)÷ 4 then 4c = −12 and solving for c we find c = −3.(c) Letting c = (−12)÷ (−4) we obtain −4c = −12 and consequently c = 3.(d) Let c = 7÷(−2). Then −2c = 7 Since there is no integer when multipliedby −2 yields 7 then we say that 7÷ (−2) is undefined in the integers

The previous example suggests the following rules of division of integers.(i) The quotient, if it exists, of two integers with the same sign is alwayspositive.(ii) The quotient, if it exists, of two integers with different signs is alwaysnegative.

Negative Exponents and Scientific NotationRecall that for any whole number a and n, a positive integer we have


.

We would like to extend this definition to include negative exponents. Thisis done by using ”looking for a pattern” strategy. Consider the following

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sequence of equalities where a is a non zero integer.

a3 = a · a · aa2 = a · aa1 = aa0 = 1

We see that each time the exponent is decreased by 1 the result is beingdivided by a. If this pattern continues we will get, a−1 = 1

a, a−2 = 1

a2 , etc. Ingeneral, we have the following definition.Let a 6= 0 be any integer and n be a positive integer then

a−n =1

an.

It can be shown that the theorems on whole-number exponents discussedin Section 10 can be extended to integer exponents. We summarize theseproperties in the following theorem.

Theorem 26.7Let a, b be integers and m,n be positive integers. Then(a) am · an = am+n

(b) am

an = am−n

(c) (am)n = amn

(d) (a · b)m = am · bm

(e)(

ab

)m= am

bm , assuming b 6= 0

(f)(

ab

)−m=

(ba

)m.

Example 26.3Write each of the following in simplest form using positive exponents in thefinal answer.(a) 162 · 8−3 (b) 202 ÷ 24 (c) (10−1 + 5 · 10−2 + 3 · 10−3) · 103

Solution.(a) 162 · 8−3 = (24)2 · (23)−3 = 28 · 2−9 = 28−9 = 2−1 = 1

2.

(b) 202 ÷ 24 = (5 · 22)2 ÷ 24 = (52 · (22)2)÷ 24 = (25 · 24)÷ 24 = 52 = 25.(c)

(10−1 + 5 · 10−2 + 3 · 10−3) · 103 = 10−1 · 103 + 5 · 10−2 · 103 + 3 · 10−3 · 103

= 10−1+3 + 5 · 10−2+3 + 3 · 10−3+3

= 102 + 5 · 10 + 3= 153

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In application problems that involve very large or very small numbers, weuse scientific notation to represent these numbers. A number is said to be inscientific notation when it is expressed in the form a×10n where 1 ≤ a < 10is called the mantissa and n is an integer called the characteristic. Forexample, the diameter of Jupiter in standard notation is 143,800,000 meters.Using scientific notation this can be written in the form 1.438× 108 meters.When converting from standard notation to scientific notation and vice versarecall the following rules of multiplying by powers of 10: When you multiplyby 10n move the decimal point n positions to the right. When you multiplyby 10−n move the decimal point n positions to the left.

Example 26.4Convert as indicated.(a) 38,500,000 to scientific notation(b) 4.135× 1011 to standard notation(c) 7.2× 10−14 to standard notation(d) 0.0000961 to scientific notation.

Solution.(a) 38, 500, 000 = 3.85× 107

(b) 4.135× 1011 = 413, 500, 000, 000(c) 7.2× 10−14 = 0.000000000000072(d) 0.0000961 = 9.61× 10−5

Order of Operations on IntegersWhen addition, subtraction, multiplication, division, and exponentiation ap-pear without parentheses, exponentiation is done first, then multiplicationand division are done from left to right, and finally addition and subtractionfrom left to right. Arithmetic operations that appear inside parentheses mustbe done first.

Example 26.5Evaluate each of the following.(a) (2− 5) · 4 + 3(b) 2 + 16÷ 4 · 2 + 8(c) 2− 3 · 4 + 5 · 2− 1 + 5.

Solution.(a) (2− 5) · 4 + 3 = (−3) · 4 + 3 = −12 + 3 = −9.

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(b) 2 + 16÷ 4 · 2 + 8 = 2 + 4 · 2 + 8 = 2 + 8 + 8 = 18.(c) 2− 3 · 4 + 5 · 2− 1 + 5 = 2− 12 + 10− 1 + 5 = 4.

Practice Problems

Problem 26.20Perform these divisions.(a) 36÷9 (b) (−36)÷9 (c) 36÷ (−9) (d) (−36)÷ (−9) (e) 165÷ (−11)(f) 275÷ 11

Problem 26.21Write another multiplication equation and two division equations that areequivalent to the equation

(−11) · (−25, 753) = 283, 283.

Problem 26.22Write two multiplication equations and another division equation that areequivalent to the equation

(−1001)÷ 11 = −91.

Problem 26.23Use the definition of division to find each quotient, if possible. If a quotientis not defined, explain why.(a) (−40)÷ (−8) (b) (−143)÷ 11 (c) 0÷ (−5) (d) (−5)÷ 0 (e) 0÷ 0

Problem 26.24Find all integers x (if possible) that make each of the following true.(a) x ÷ 3 = −12 (b) x ÷ (−3) = −2 (c) x ÷ (−x) = −1 (d) 0 ÷ x = 0(e) x÷ 0 = 1

Problem 26.25Write two division equations that are equivalent to 3× (−2) = −6.

Problem 26.26Explain how to compute −10÷ 2 using signed counters.

28

Problem 26.27Rewrite each of the following as an equivalent multiplication problem, andgive the solution.(a) (−54)÷ (−6) (b) 32÷ (−4)

Problem 26.28A store lost $480,000 last year.(a) What was the average net change per month?(b) Write an integer equation for this situation.

Problem 26.29Compute the following, using the correct rules for order of operations.(a) −22 − 3 (b) −5 + (−4)2 × (−2)

Problem 26.30A stock change as follows for 5 days:-2,4,6,3,-1. What is the average dailychange in price?

Problem 26.31Compute: −2÷ (−2) + (−2)− (−2)

Problem 26.32For what integers a and b does a÷ b = b÷ a?

Problem 26.33Find each quotient, if possible.(a) [144÷ (−12)]÷ (−3) (b) 144÷ [−12÷ (−3)]

Problem 26.34Compute the following writing the final answer in terms of positive exponents.(a) 4−2 · 46

(b) 63

6−7

(c) (3−4)−2

Problem 26.35Express each of the following in scientific notation.(a) 0.0004 (b) 0.0000016 (c) 0.000000000000071

29

Problem 26.36Hair on the human body can grow as fast as 0.0000000043 meter per second.(a) At this rate, how much would a strand of hair grow in one month of 30days? Express your answer in scientific notation.(b) About how long would it take for a strand of hair to grow to be 1 meterin length?

Problem 26.37Compute each of these to three significant figures using scientific notation.(a) (2.47× 10−5) · (8.15× 10−9)(b) (2.47× 10−5)÷ (8.15× 10−9)

Problem 26.38Convert each of the following to standard notation.(a) 6.84× 10−5 (b) 3.12× 107

Problem 26.39Write each of the following in scientific notation.(a) 413,682,000 (b) 0.000000231 (c) 100,000,000

Problem 26.40Evaluate each of the following.(a) −52 + 3(−2)2

(b) −2 + 3 · 5− 1(c) 10− 3 · 7− 4(−2)÷ 2 + 3

Problem 26.41Evaluate each of the following, if possible.(a) (−10÷ (−2))(−2)(b) (−10 · 5)÷ 5(c) −8÷ (−8 + 8)(d) (−6 + 6)÷ (−2 + 2)(e) | − 24| ÷ 4 · (3− 15)

Comparing and Ordering IntegersIn this section we extend the notion of ”less than” to the set of all integers.We describe two equivalent ways for viewing the meaning of less than: a

30

number line approach and an addition (or algebraic) approach. In what fol-lows, a and b denote any two integers.

Number-Line ApproachWe say that a is less than b, and we write a < b, if the point representinga on the number-line is to the left of b. For example, Figure 26.3 shows that−4 < 2.

Figure 26.3

Example 26.6Order the integers −1, 0, 4, 3,−2, 2,−4,−3, 1 from smallest to largest usingthe number line approach.

Solution.The given numbers are ordered as shown in Figure 26.4

Figure 26.4

Addition ApproachNote that from Figure 26.3 we have 2 = (−4) + 6 so we say that 2 is 6 morethan −2. In general, if a < b then we can find a unique integer c such thatb = a + c.

Example 26.7Use the addition approach to show that −7 < −3.

Solution.Since −3 = (−7) + 4 then −3 is 4 more than −7, that is −7 < −3

Notions similar to less than are included in the following table.

31

Inequality Symbol Meaning< less than> greater than≤ less than or equal≥ greater than or equal

The following rules are valid for any of the inequality listed in the above table.

Rules for Inequalities• Trichotomy Law: For any integers a and b exactly one of the followingis true:

a < b, a > b, a = b.

• Transitivity: For any integers a, b, and c if a < b and b < c then a < c.•Addition Property: For any integers a, b, and c if a < b then a+c < b+c.• Multiplication Property: For any integers a, b, and c if a < b thenac < bc if c > 0 and ac > bc if c < 0.

The first three properties are extensions of similar statements in the wholenumber system. Note that the fourth property asserts that an inequalitysymbol must be reversed when multiplying by a negative integer. This isillustrated in Figure 26.5 when c = −1.

Figure 26.5

Practice Problems

Problem 26.42Use the number-line approach to verify each of the following.(a) −4 < 1 (b) −4 < −2 (c) −1 > −5

Problem 26.43Order each of the following lists from smallest to largest.(a) {−4, 4,−1, 1, 0}(b) {23,−36, 45,−72,−108}

32

Problem 26.44Replace the blank by the appropriate symbol.(a) If x > 2 then x + 4 6(b) If x < −3 then x− 6 − 9

Problem 26.45Determine whether each of the following statements is correct.(a) −3 < 5 (b) 6 < 0 (c) 3 ≤ 3 (d) −6 > −5 (e) 2× 4− 6 ≤ −3× 5 + 1

Problem 26.46What different looking inequality means the same as a < b?

Problem 26.47Use symbols of inequalities to represent ”at most” and ”at least”.

Problem 26.48For each inequality, determine which of the numbers −5, 0, 5 satisfies theinequality.(a) x > −5 (b) 5 < x (c) −5 > x

Problem 26.49Write the appropriate inequality symbol in the blank so that the two inequal-ities are true.(a) If x ≤ −3 then −2x 6(b) If x + 3 > 9 then x 6

Problem 26.50How do you know when to reverse the direction of an inequality symbol?

Problem 26.51Show that each of the following inequality is true by using the additionapproach.(a) −4 < −2 (b) −5 < 3 (c) −17 > −23

Problem 26.52A student makes a connection between debts and negative numbers. So thenumber −2 represents a debt of $2. Since a debt of $10 is larger than a debtof $5 then the student writes −10 > −5. How would you convince him/herthat this inequality is false?

33

Problem 26.53At an 8% sales tax rate, Susan paid more than $1500 sales tax when shepurchased her new Camaro. Describe this situation using an inequality withp denoting the price of the car.

Problem 26.54Show that if a and b are positive integers with a < b then a2 < b2. Does thisresult hold for any integers?

Problem 26.55Elka is planning a rectangular garden that is twice as long as it is wide. Ifshe can afford to buy at most 180 feet of fencing, then what are the possiblevalues for the width?

34

27 Rational Numbers

Integers such as −5 were important when solving the equation x+5 = 0. In asimilar way, fractions are important for solving equations like 2x = 1. Whatabout equations like 2x + 1 = 0? Equations of this type require numbers like−1

2. In general, numbers of the form a

bwhere a and b are integers with b 6= 0

are solutions to the equation bx = a. The set of all such numbers is the setof rational numbers, denoted by Q :

Q = {a

b: a, b ∈ Z, b 6= 0}.

That is, the set of rational numbers consists of all fractions with their oppo-sites. In the notation a

bwe call a the numerator and b the denominator.

Note that every fraction is a rational number. Also, every integer is a rationalnumber for if a is an integer then we can write a = a

1. Thus, Z ⊂ Q.

Example 27.1Draw a Venn diagram to show the relationship between counting numbers,whole numbers, integers, and rational numbers.

Solution.The relationship is shown in Figure 27.1

Figure 27.1

All properties that hold for fractions apply as well for rational numbers.

Equality of Rational Numbers: Let ab

and cd

be any two rational numbers.Then a

b= c

dif and only if ad = bc (Cross-multiplication).

35

Example 27.2Determine if the following pairs are equal.(a) 3

−12and −36

144.

(b) −2186

and −51215

.

Solution.(a) Since 3(144) = (−12)(−36) then 3

−12= −36

144.

(b) Since (−21)(215) 6= (86)(−51) then −21866= −51

215

The Fundamental Law of Fractions: Let ab

be any rational numberand n be a nonzero integer then

a

b=

an

bn=

a÷ n

b÷ n.

As an important application of the Fundamental Law of Fractions we have

a

−b=

(−1)a

(−1)(−b)=−a

b.

We also use the notation −ab

for either a−b

or −ab

.

Example 27.3Write three rational numbers equal to −2

5.

Solution.By the Fundamental Law of Fractions we have

−2

5=

4

−10= − 6

15=−8

20

Rational Numbers in Simplest Form: A rational number ab

is in sim-plest form if a and b have no common factors greater than 1.The methods of reducing fractions into simplest form apply as well withrational numbers.

Example 27.4Find the simplest form of the rational number 294

−84.

Solution.Using the prime factorizations of 294 and 84 we find

294

−84=

2 · 3 · 72

(−2) · 2 · 3 · 7=

7

−2=−2

7

36

Practice Problems

Problem 27.1Show that each of the following numbers is a rational number.(a) −3 (b) 41

2(c) −5.6 (d) 25%

Problem 27.2Which of the following are equal to −3?

−3

1,

3

−1,3

1,−3

1,−3

−1,−−3

1,−−3

−1.

Problem 27.3Determine which of the following pairs of rational numbers are equal.(a) −3

5and 63

−105

(b) −18−24

and 4560

.

Problem 27.4Rewrite each of the following rational numbers in simplest form.(a) 5

−7(b) 21

−35(c) −8

−20(d) −144

180

Problem 27.5How many different rational numbers are given in the following list?

2

5, 3,

−4

−10,39

13,7

4

Problem 27.6Find the value of x to make the statement a true one.

(a) −725

= x500

(b) 183

= −5x

Problem 27.7Find the prime factorizations of the numerator and the denominator and usethem to express the fraction 247

−77in simplest form.

Problem 27.8(a) If a

b= a

c, what must be true?

(b) If ac

= bc, what must be true?

37

Addition of Rational NumbersThe definition of adding fractions extends to rational numbers.

a

b+

c

b=

a + c

b

a

b+

c

d=

ad + bc

bd

Example 27.5Find each of the following sums.(a) 2

−3+ 1

5(b) −2

5+ 4

−7(c) 3

7+ −5

7

Solution.(a) 2

−3+ 1

5= 2·(−5)

(−3)·(−5)+ 1·3

5·3 = −1015

+ 315

= (−10)+315

= −715

(b) −25

+ 4−7

= (−2)·75·7 + 4·(−5)

(−5)·(−7)= −14

35+ −20

35= (−14)+(−20)

35= −34

35.

(c) 37

+ −57

= 3+(−5)7

= −27

Rational numbers have the following properties for addition.

Theorem 27.1Closure: a

b+ c

dis a unique rational number.

Commutative: ab

+ cd

= cd

+ ab

Associative:(

ab

+ cd

)+ e

f= a

b+

(cd

+ ef

)Identity Element: a

b+ 0 = a

b

Additive inverse: ab

+(−a

b

)= 0

Example 27.6Find the additive inverse for each of the following:(a) 3

5(b) −5

11(c) 2

−3(d) −2

5

Solution.(a) −3

5= −3

5= 3

−5

(b) 511

(c) 23

(d) 25

38

Subtraction of Rational NumbersSubtraction of rational numbers like subtraction of fractions can be definedin terms of addition as follows.

a

b− c

d=

a

b+

(− c

d

).

Using the above result we obtain the following:

ab− c

d= a

b+

(− c

d

)= a

b+ −c

d

= ad+b(−c)bd

= ad−bcbd

Example 27.7Compute 103

24− −35

16.

Solution.

10324− −35

16= 206

48− −105

48

= 206−(−105)48

= 206+10548

= 31148

Practice Problems

Problem 27.9Use number line model to illustrate each of the following sums.(a) 3

4+ −2

4(b) −3

4+ 2

4(c) −3

4+ −1

4

Problem 27.10Perform the following additions. Express your answer in simplest form.(a) 6

8− −25

100(b) −57

100+ 13

10

Problem 27.11Perform the following subtractions. Express your answer in simplest form.(a) 137

214− −1

3(b) −23

100− 198

1000

Problem 27.12Compute the following differences.(a) 2

3− −9

8(b)

(−21

4

)− 42

3

39

Multiplication of Rational NumbersThe multiplication of fractions is extended to rational numbers. That is, ifab

and cd

are any two rational numbers then

a

b· c

d=

ac

bd.

Multiplication of rational numbers has properties analogous to the proper-ties of multiplication of fractions. These properties are summarized in thefollowing theorem.

Theorem 27.2Let a

b, c

d, and e

fbe any rational numbers. Then we have the following:

Closure: The product of two rational numbers is a unique rational number.Commutativity: a

b· c

d= c

d· a

b.

Associativity: ab·(

cd· e

f

)=

(ab· c

d

)· e

f.

Identity: ab· 1 = a

b= 1 · a

b.

Inverse: ab· b

a= 1. We call b

athe reciprocal of a

bor the multiplicative

inverse of ab.

Distributivity: ab·(

cd

+ ef

)= a

b· c

d+ a

b· e

f.

Example 27.8Perform each of the following multiplications. Express your answer in sim-plest form.

(a) −56· 7

3(b) −3

10· −25

27

Solution.(a) We have

−5

6· 7

3=

(−5) · 76 · 3

=−35

18.

(b)−3

10· −25

27=−1

2· −5

9=

(−1)(−5)

2(9)=

5

18

Example 27.9Use the properties of multiplication of rational numbers to compute the fol-lowing.

40

(a) −35·(

1117· 5

3

)(b) 2

3·(

32

+ 57

)(c) 5

9· 2

7+ 2

7· 4

9

Solution.(a) −3

5·(

1117· 5

3

)= −3

5· 11·5

17·3 = −35· 55

51= −1

1· 11

17= −11

17

(b) 23·(

32

+ 57

)= 2

3· 31

14= 1

3· 31

7= 31

21

(c) 59· 2

7+ 2

7· 4

9= 5

9· 2

7+ 4

9· 2

7=

(59

+ 49

)· 2

7= 2

7

Division of Rational NumbersWe define the division of rational numbers as an extension of the division offractions. Let a

band c

dbe any rational numbers with c

d6= 0. Then

a

b÷ c

d=

a

b· d

c.

Using words, to find ab÷ c

dmultiply a

bby the reciprocal of c

d.

By the above definition one gets the following two results.

a

b÷ c

b=

a

c

anda

b÷ c

d=

a÷ c

b÷ d.

Remark 27.1After inverting, it is often simplest to ”cancel” before doing the multiplica-tion. Cancelling is dividing one factor of the numerator and one factor of thedenominator by the same number. For example: 2

9÷ 3

12= 2

9× 12

3= 2×12

9×3=

2×43×3

= 89.

Remark 27.2Exponents and their properties are extended to rational numbers in a naturalway. For example, if a is any rational number and n is a positive integer then

an = a · a · · · a︸︷︷︸n factors

and a−n =1

an

41

Example 27.10Compute the following and express the answers in simplest form.

(a) −74÷ 2

3(b) 13

17· −4

9(c) −18

23· −6

23

Solution.(a) −7

4÷ 2

3= −7

4· 3

2= (−7)(3)

(4)(2)= −21

8.

(b) 1317· −4

9= 13

17· 9−4

= 13·917·(−4)

= −11768

.

(c) −1823· −6

23= −18

23· 23−6

= 31· 1

1= 3

Practice Problems

Problem 27.13Multiply the following rational numbers. Write your answers in simplestform.

(a) 35· −10

21(b) −6

11· −33

18(c) 5

12· 48−15

· −98

Problem 27.14Find the following quotients. Write your answers in simplest form.

(a) −89÷ 2

9(b) 12

15÷ −4

3(c) −13

24÷ −39

−48

Problem 27.15State the property that justifies each statement.

(a)(

57· 7

8

)· −8

3= 5

7·(

78· −8

3

)(b) 1

4

(83

+ −54

)= 1

4· 8

3+ 1

4· −5

4

Problem 27.16Compute the following and write your answers in simplest form.

(a) −4027÷ −10

9(b) 21

25÷ −3

5(c) −10

9÷ −9

8

Problem 27.17Find the reciprocals of the following rational numbers.

(a) 4−9

(b) 0 (c) −32

(d) −4−9

42

Problem 27.18Compute:

(−47· 2−5

)÷ 2

−7.

Problem 27.19If a

b· −4

7= 2

3what is a

b?

Problem 27.20Compute −41

2×−52

3


9÷ 510

11

Problem 27.22Compute each of the following:

(a) −(

34

)2(b)

(−3

4

)2(c)

(34

)2 ·(

34

)7

Comparing and Ordering Rational NumbersIn this section we extend the notion of ”less than” to the set of all ratio-nals. We describe two equivalent ways for viewing the meaning of less than:a number line approach and an addition (or algebraic) approach. In whatfollows, a

band c

ddenote any two rationals.

Number-Line ApproachWe say that a

bis less than c

d, and we write a

b< c

d, if the point representing

ab

on the number-line is to the left of cd. For example, Figure 27.2 shows that

12

< 23.

Figure 27.2

Example 27.11Use the number line approach to order the pair of numbers 3

7and 5

2.

Solution.When the two numbers have unlike denominators then we find the least com-mon denominator and then we order the numbers. Thus, 3

7= 6

14and 5

2= 35

14.

43

Hence, on a number line, 37

is to the left of 52.

Addition ApproachAs in the case of ordering integers, we say that a

b< c

dif there is a unique

fraction ef

such that ab

+ ef

= cd.

Example 27.12Use the addition approach to show that −3

7< 5

2.

Solution.Since 5

2= −3

7+ 41

14then −3

7< 5

2

Notions similar to less than are included in the following table.

Inequality Symbol Meaning< less than> greater than≤ less than or equal≥ greater than or equal

The following rules are valid for any of the inequality listed in the above table.

Rules for Inequalities• Trichotomy Law: For any rationals a

band c

dexactly one of the following

is true:a

b<

c

d,a

b>

c

d,a

b=

c

d.

• Transitivity: For any rationals ab, c

d, and e

fif a

b< c

dand c

d< e

fthen a

b< e

f.

• Addition Property: For any rationals ab, c

d, and e

fif a

b< c

dthen a

b+ e

f<

cd

+ ef.

• Multiplication Property: For any rationals ab, c

d, and e

fif a

b< c

dthen

ab· e

f< c

d· e

fif e

f> 0 and a

b· e

f> c

d· e

fif e

f< 0

• Density Property: For any rationals ab

and cd, if a

b< c

dthen

a

b<

1

2

(a

b+

c

d

)<

c

d.

Practice Problems

Problem 27.23True or false: (a) −2

3< −3

7(b) 15

−9> −13

4.

44

Problem 27.24Show that −3

4< −1

4using the addition approach.


4< −1

4by using a number line.

Problem 27.26Put the appropriate symbol, <, =, > between each pair of numbers to makea true statement.

(a) −56

−1112

(b) −13

54

(c) −1215

36−45

(d) − 312

−420

Problem 27.27Find three rational numbers between 1

4and 2

5.

Problem 27.28The properties of rational numbers are used to solve inequalities. For exam-ple,

x + 35

< −710

x + 35

+(−3

5

)< −7

10+

(−3

5

)x < −13

10

Solve the inequality

−2

5x +

1

5> −1.

Problem 27.29Solve each of the following inequalities.

(a) x− 65

< −127

(b) 25x < −7

8

(c) −37

x > 85

45

Problem 27.30Verify the following inequalities.

(a) −45

< −34

(b) 110

< 14

(c) 19−60

> −13

Problem 27.31Use the number-line approach to arrange the following rational numbers inincreasing order:

(a) 45,−1

5, 2

5

(b) −712

, −23

, 3−4

Problem 27.32Find a rational number between 5

12and 3

8.

Problem 27.33Complete the following, and name the property that is used as a justification.

(a) If −23

< 34

and 34

< 75

then −23

75.

(b) If −35

< −611

then(−3

5

)·(

23

) (−611

)·(

23

)(c) If −4

7< 7

4then −4

7+ 5

8< 7

4+

(d) If −34

< 113

then(−3

4

)·(−5

7

) (113

)·(−5

7

)(e) There is a rational number any two unequal rational numbers.

46

28 Real Numbers

In the previous section we introduced the set of rational numbers. We haveseen that integers and fractions are rational numbers. Now, what aboutdecimal numbers? To answer this question we first mention that a decimalnumber can be terminating, nonterminating and repeating, nonterminatingand nonrepeating.If the number is terminating then we can use properties of powers of 10 towrite the number as a rational number. For example, −0.123 = − 123

1000and

15.34 = 1534100

. Thus, every terminating decimal number is a rational number.Now, suppose that the decimal number is nonterminating and repeating. Forthe sake of argument let’s take the number 12.341341 · · · where 341 repeatsindefinitely. Then 12.341341 · · · = 12 + 0.341341 · · · . Let x = 0.341341 · · · .Then 1000x = 341.341341 · · · = 341 + x. Thus, 999x = 341 and thereforex = 341

999. It follows that

12.341341 · · · = 12 +341

999=

12329

999

Hence, every nonterminating repeating decimal is a rational number.Next, what about nonterminating and non repeating decimals. Such numbersare not rationals and the collection of all such numbers is called the set ofirrational numbers. As an example of irrational numbers, let’s considerfinding the hypotenuse c of a right triangle where each leg has length 1. Thenby the Pythagorean formula we have

c2 = 12 + 12 = 2.

We will show that c is irrational. Suppose the contrary. That is, supposethat c is rational so that it can be written as

a

b= c

where a and b 6= 0 are integers. Squaring both sides we find a2

b2= c2 = 2 or

a2 = 2 · b2. If a has an even number of prime factors then a2 has an evennumber of prime factors. If a has an odd number of prime factors then a2

has an even number of prime factors. So, a2 and b2 have both even numberof prime factors. But 2 · b2 has an odd number of prime factors. So wehave that a2 has an even number and an odd number of prime factors. This

47

cannot happen by the Fundamental Theorem of Arithmetic which says thatevery positive integer has a unique number of prime factors. In conclusion,c cannot be written in the form a

bso it is not rational. Hence, its decimal

form is nonterminating and nonrepeating.

Remark 28.1It follows from the above discussion that the equation c2 = 2 has no answersin the set of rationals. This is true for the equation c2 = p where p is a primenumber. Since there are infinite numbers of primes then the set of irrationalnumbers is infinite.

We define the set of real numbers to be the union of the set of rationalsand the set of irrationals. We denote it by the letter R. The relationshipsamong the sets of all numbers discussed in this book is summarized in Figure28.1

Figure 28.1

Now, in the set of real numbers, the equation c2 = 2 has a solution denotedby√

2. Thus, (√

2)2 = 2. In general, we say that a positive number b is thesquare root of a positive number a if b2 = a. We write

√a = b.

Representing Irrational Numbers on a Number LineFigure 28.2 illustrates how we can accurately plot the length

√2 on the

number line.

Figure 28.2

48

Practice Problems

Problem 28.1Given a decimal number, how can you tell whether the number is rational orirrational?

Problem 28.2Write each of the following repeating decimal numbers as a fraction.(a) 0.8 (b) 0.37 (c) 0.02714

Problem 28.3Which of the following describe 2.6?(a) A whole number(b) An integer(c) A rational number(d) An irrational number(e) A real number

Problem 28.4Classify the following numbers as rational or irrational.(a)

√11 (b) 3

7(c) π (d)

√16

Problem 28.5Classify the following numbers as rational or irrational.(a) 0.34938661 · · · (b) 0.26 (c) 0.565665666 · · ·

Problem 28.6Match each word in column A with a word in column B.

A BTerminating RationalRepeating IrrationalInfinite nonrepeating

Problem 28.7(a) How many whole numbers are there between −3 and 3 (not including 3and −3)?(b) How many integers are there between −3 and 3?(c) How many real numbers are there between −3 and 3?

49

Problem 28.8Prove that

√3 is irrational.

Problem 28.9Show that 1 +

√3 is irrational.

Problem 28.10Find an irrational number between 0.37 and 0.38

Problem 28.11Write an irrational number whose digits are twos and threes.

Problem 28.12Classify each of the following statement as true or false. If false, give acounter example.

(a) The sum of any rational number and any irrational number is a rationalnumber.(b) The sum of any two irrational numbers is an irrational number.(c) The product of any two irrational numbers is an irrational number.(d) The difference of any two irrational numbers is an irrational number.

Arithmetic of Real NumbersAddition, subtraction, multiplication, division, and ”less than” are definedon the set of real numbers in such a way that all the properties on the ratio-nals still hold. These properties are summarized next.

Closure: For any real numbers a and b, a + b and ab are unique real num-bers.Commutative: For any real numbers a and b, a + b = b + a and ab = ba.Associative: For any real numbers, a, b, and c we have a+(b+c) = (a+b)+cand a(bc) = (ab)c.Identity element: For any real numbers a we have a + 0 = a and a · 1 = a.Inverse Elements: For any real numbers a and b 6= 0 we have a+(−a) = 0and b · 1

b= 1.

Distributive: For any real numbers a, b, and c we have a(b + c) = ab + ac.Transitivity: If a < b and b < c then a < c.Addition Property: If a < b then a + c < b + c.Multiplication Property: If a < b then ac < bc if c > 0 and ac > bc if

50

c < 0.Density: If a < b then a < a+b

2< b.

Practice Problems

Problem 28.13Construct the lengths

√2,√

3,√

4,√

5, · · · as follows.(a) First construct a right triangle with both legs of length 1. What is thelength of the hypotenuse?(b) This hypotenuse is the leg of the next right triangle. The other leg haslength 1. What is the length of the hypotenuse of this triangle?(c) Continue drawing right triangles, using the hypotenuse of the proceedingtriangle as a leg of the next triangle until you have constructed one withlength

√6.

Problem 28.14Arrange the following real numbers in increasing order.0.56, 0.56, 0.566, 0.56565556 · · ·, 0.566

Problem 28.15Which property of real numbers justify the following statement

2√

3 + 5√

3 = (2 + 5)√

3 = 7√

3

Problem 28.16Find x : x + 2

√2 = 5

√2.

Problem 28.17Solve the following equation: 2x− 3

√6 = 3x +

√6

Problem 28.18Solve the inequality: 3

2x− 2 < 5

6x + 1

3

Problem 28.19Determine for what real number x, if any, each of the following is true:(a)

√x = 8 (b)

√x = −8 (c)

√−x = 8 (d)

√−x = −8

Radical and Rational ExponentsBy rational exponents we mean exponents of the form

amn

where a is any real number and m and n are integers. First we consider thesimple case a

1n (where n is a positive integer) which is defined as follows

51

a1n = b is equivalent to bn = a.

We call b the nth root of a and we write b = n√

a. We call a the radicandand n the index. The symbol n

√is called a radical. It follows that

a1n = n

√a.

Note that the above definition requires a ≥ 0 for n even since bn is alwaysgreater than or equal to 0.

Example 28.1Compute each of the following.(a) 25

12 (b) (−8)

13 (c) (−16)

14 (d) −16

14

Solution.(a) Since 52 = 25 then 25

12 = 5.

(b) Since (−2)3 = −8 then (−8)13 = −2.

(c) Since the radicand is negative and the index is even then (−16)14 is un-

defined.(d) Since 24 = 16 then −16

14 = −2

As the last two parts of the previous example have once again shown, wereally need to be careful with parentheses. In this case parenthesis makesthe difference between being able to get an answer or not.

For a negative exponent we define

a−1n =

1

a1n

.

Now, for a general rational exponent we define

amn =

(a

1n

)m

.

That is,a

mn = ( n

√a)m.

Now, if b = (a1n )m then bn = ((a

1n )m)n = (a

1n )mn = ((a

1n )n)m = (a

nn )m = am.

This shows that b = (am)1n . Hence, we have established that

(a1n )m = (am)

1n

or( n√

a)m = n√

am.

52

Example 28.2Rewrite each expression using rational exponents.

(a) 4√

5xy (b)3√

4a2b5

Solution.(a) 4

√5xy = (5xy)

14

(b)3√

4a2b5 = (4a2b5)13

Example 28.3Express the following values without exponents.

(a) 932 (b) 125−

43 .

Solution.(a) 9

32 = (9

12 )3 = 33 = 27.

(b) 125−43 = (125

13 )−4 = 5−4 = 1

625.

The properties of integer exponents also hold for rational exponents. Theseproperties are equivalent to the corresponding properties of radicals if theexpressions involving radicals are meaningful.

Theorem 28.1Let a and b real numbers and n a nonzero integer. Then

(ab)1n = a

1n b

1n

n√

ab = n√

a n√

b(ab

) 1n = a

1n

b1n

n√

ab

=n√an√

b

Practice Problems

Problem 28.20Express the following values without exponents.(a)25

12 (b) 32

15 (c) 9

52 (d) (−27)

43

Problem 28.21Write the following radicals in simplest form if they are real numbers.(a) 3

√−27 (b) 4

√−16 (c) 5

√32

53

Problem 28.22A student uses the formula

√a√

b =√

ab to show that -1 = 1 as follows:

−1 = (√−1)2 =

√−1√−1 =

√(−1)(−1) =

√1 = 1.

What’s wrong with this argument?

Problem 28.23Give an example where the following statement is true:

√a + b =

√a +

√b

Problem 28.24Give an example where the following statement is false:

√a + b =

√a +

√b.

Problem 28.25Find an example where the following statement is false:

√a2 + b2 = a + b.

Problem 28.26Express the following values without using exponents:(a) (310)

35 (b) 81−

54

Problem 28.27If possible, find the square root of each of the following without using acalculator.(a) 225 (b) 169 (c) -81 (d) 625

Problem 28.28Write each of the following in the form a

√b where a and b are integers and

b has the least value possible.(a)

√180 (b)

√363 (c)

√252

54

29 Functions and their Graphs

The concept of a function was introduced and studied in Section 7 of thesenotes. In this section we explore the graphs of functions. Of particular in-terest, we consider the graphs of linear functions, quadratic functions, cubicfunctions, square root functions, and exponential functions. These graphsare represented in a coordinate system known as the Cartesian coordi-nate system which we explore next.

The Cartesian PlaneThe Cartesian coordinate system was developed by the mathematician ReneDescartes in 1637. The Cartesian coordinate system, also known asthe rectangular coordinate system or the xy-plane, consists of two numberscales, called the x-axis and the y-axis, that are perpendicular to each otherat point O called the origin. Any point in the system is associated with anordered pair of numbers (x, y) called the coordinates of the point. Thenumber x is called the abscissa or the x-coordinate and the number y iscalled the ordinate or the y-coordinate. The abscissa measures the dis-tance from the point to the y-axis whereas the ordinate measures the distanceof the point to the x-axis. Positive values of the x-coordinate are measuredto the right, negative values to the left. Positive values of the y-coordinateare measured up, negative values down. The origin is denoted as (0, 0).The axes divide the coordinate system into four regions called quadrantsand are numbered counterclockwise as shown in Figure 29.1To plot a point P (a, b) means to draw a dot at its location in the xy-plane.

Example 29.1Plot the point P with coordinates (5, 2).

Solution.Figure 29.1 shows the location of the point P (5, 2) in the xy-plane.

55

Figure 29.1

Example 29.2Complete the following table of signs of the coordinates of a point P (x, y).

x yQuadrant IQuadrant IIQuadrant IIIQuadrant IVPositive x-axisNegative x-axisPositive y-axisNegative y-axis

56

Solution.

x yQuadrant I + +Quadrant II - +Quadrant III - -Quadrant IV + -Positive x-axis + 0Negative x-axis - 0Positive y-axis 0 +Negative y-axis 0 -

When Does a Graph Represent a FunctionTo say that y is a function of x means that for each value of x there is exactlyone value of y. Graphically, this means that each vertical line must intersectthe graph at most once. Hence, to determine if a graph represents a functionone uses the following simple visual test:

Vertical Line Test: A graph is a function if and only if every verticalline crosses the graph at most once.

According to the vertical line test and the definition of a function, if a ver-tical line cuts the graph more than once, the graph could not be the graphof a function since we have multiple y values for the same x-value and thisviolates the definition of a function.

Example 29.3Which of the graphs (a), (b), (c) in Figure 29.2 represent y as a function of x?

57

Figure 29.2

Solution.By the vertical line test, (b) represents a function whereas (a) and (c) fail torepresent functions since one can find a vertical line that intersects the graphmore than once.

The domain of a function is the collection of all possible x-coordinates thatcan be used in the formula of the function. For example, x = 1 is in thedomain of f(x) = x + 1 since f(1) = 1 + 1 = 2 whereas x = 1 is not in thedomain of f(x) = 1

x−1since f(1) = 1

0which is undefined.

The collection of all values of y-coordinates that correspond to the x-coordinatesis called the range of the function. For example, the range of f(x) =

√x− 1

is the interval [0,∞) whereas that of the function f(x) = 1x−1

is the setR− {0}.

Practice Problems

Problem 29.1Plot the points whose coordinates are given on a Cartesian coordinate system.

(a) (2, 4), (0,−3), (−2, 1), (−5,−3).(b) (−3,−5), (−4, 3), (0, 2), (−2, 0).

58

Problem 29.2Plot the following points using graph papers.(a) (3,2) (b) (5,0) (c) (0,-3) (d) (-3,4) (e) (-2,-3) (f) (2,-3)

Problem 29.3Complete the following table.

(x,y) x > 0, y > 0 x < 0, y > 0 x > 0, y < 0 x < 0, y < 0Quadrant

Problem 29.4In the Cartesian plane, shade the region consisting of all points (x, y) thatsatisfy the two conditions

−3 ≤ x ≤ 2 and 2 ≤ y ≤ 4

Problem 29.5Determine which of the following graphs represent a function.

Problem 29.6Consider the function f whose graph is given below.

59

(a) Complete the following table

x -2 -1 0 1 2 3f(x)

(b) Find the domain and range of f.(c) For which values of x is f(x) = 2.5?

Graphs of Linear FunctionsA linear function is any function that can be written in the form f(x) =mx+ b. As the name suggests, the graph of such a function is a straight line.

Example 29.4The sales tax on an item is 6%. So if p denotes the price of the item and Cthe total cost of buying the item then if the item is sold at $ 1 then the costis 1 + (0.06)(1) = $1.06 or C(1) = $1.06. If the item is sold at $2 then thecost of buying the item is 2 + (0.06)(2) = $2.12, or C(2) = $2.12, and so on.Thus we have a relationship between the quantities C and p such that eachvalue of p determines exactly one value of C. In this case, we say that C is afunction of p. Find a formula for p and graph.

Solution.The chart below gives the total cost of buying an item at price p as a functionof p for 1 ≤ p ≤ 6.

p 1 2 3 4 5 6C 1.06 2.12 3.18 4.24 5.30 6.36

60

The graph of the function C is obtained by plotting the data in the abovetable. See Figure 29.3.The formula that describes the relationship between C and p is given by

C(p) = 1.06p.

Figure 29.3

Graphs of Quadratic FunctionsYou recall that a linear function is a function that involves a first power ofx. A function of the form

f(x) = ax2 + bx + c, a 6= 0

is called a quadratic function. The word ”quadratus” is the latin word fora square.Quadratic functions are useful in many applications in mathematics when alinear function is not sufficient. For example, the motion of an object throwneither upward or downward is modeled by a quadratic function.The graph of a quadratic function is a curve called a parabola. Parabolasmay open upward or downward and vary in ”width” or ”steepness”, but theyall have the same basic ”U” shape.All parabolas are symmetric with respect to a line called the axis of sym-metry. A parabola intersects its axis of symmetry at a point called thevertex of the parabola.

61

Many quadratic functions can be graphed easily by hand using the techniquesof stretching/shrinking and shifting (translation) the parabola y = x2.

Example 29.5Sketch the graph of y = x2

2.

Solution.Starting with the graph of y = x2, we shrink by a factor of one half. Thismeans that for each point on the graph of y = x2, we draw a new point thatis one half of the way from the x-axis to that point. See Figure 29.4

Figure 29.4

When a quadratic function is in standard form, then it is easy to sketch itsgraph by reflecting, shifting, and stretching/shrinking the parabola y = x2.

The quadratic function f(x) = a(x− h)2 + k, a not equal to zero, is said tobe in standard form. If a is positive, the graph opens upward, and if a isnegative, then it opens downward. The line of symmetry is the vertical linex = h, and the vertex is the point (h, k).Any quadratic function can be rewritten in standard form by completing thesquare. Note that when a quadratic function is in standard form it is alsoeasy to find its zeros by the square root principle.

Example 29.6Write the function f(x) = x2 − 6x + 7 in standard form. Sketch the graphof f and find its zeros and vertex.

62

Solution.Using completing the square method we find

f(x) = x2 − 6x + 7= (x2 − 6x) + 7= (x2 − 6x + 9− 9) + 7 (Just square 6

2)

= (x2 − 6x + 9)− 9 + 7= (x− 3)2 − 2

From this result, one easily finds the vertex of the graph of f is (3,−2).To find the zeros of f, we set f equal to 0 and solve for x.

(x− 3)2 − 2 = 0(x− 3)2 = 2

x− 3 = ±√

2

x = 3±√

2

Finally, the graph of f is given in Figure 29.5

Figure 29.5

Graphs of Exponential FunctionsAn exponential function is a function that can be written in the form

f(t) = b · at

where a is positive and different from 1. We call a the base of the function.Figure 29.6 shows the graph of an exponential function.

63

Figure 29.6

For a > 1 the function is increasing. In this case, we say that the functionrepresents an exponential growth. If 0 < a < 1 then the function representsan exponential decay.

Remark 29.1Why a is restricted to a > 0 and a 6= 1? Since t is allowed to have any valuethen a negative a will create meaningless expressions such as

√a (if t = 1

2).

Also, for a = 1 the function P (t) = b is called a constant function and itsgraph is a horizontal line.

Miscellaneous FunctionsWe consider some of the graphs of some important functions.

Square Root FunctionThe square root function is the function f(x) =

√x. To get the graph well

just plug in some values of x and then plot the points.

x 0 1 4 9f(x) 0 1 2 3

64

The graph is given in Figure 29.7

Figure 29.7

Absolute Value FunctionWe’ve dealt with this function several times already. It’s now time to graph it.First, let’s remind ourselves of the definition of the absolute value function.

f(x) =

{x if x ≥ 0−x if x < 0

Finding some points to plot we get

x -2 -1 0 1 2f(x) 2 1 0 1 2


65

Figure 29.8

Cubic FunctionWe will consider the following simple form of a cubic function f(x) = x3.First, we find some points on the graph.

x -2 -1 0 1 2f(x) -8 -1 0 1 8


Figure 29.9

Step FunctionsThe far we have considered functions whose graphs are ”continuous”, thatis, the graphs have no holes or jumps. Step functions are functions that arenot continuous.

Example 29.7The charge for a taxi ride is $1.50 for the first 1

5of a mile, and $0.25 for each

additional 15

of a mile (rounded up to the nearest 15

mile).

(a) Sketch a graph of the cost function C as a function of the distance trav-eled x, assuming that 0 ≤ x ≤ 1.(b) Find a formula for C in terms of x on the interval [0, 1].(c) What is the cost for a 4

5−mile ride?

66

Solution.(a) The graph is given in Figure 29.10.

Figure 29.10

(b) A formula of C(x) is

C(x) =

1.50 if 0 ≤ x ≤ 1

5

1.75 if 15

< x ≤ 25

2.00 if 25

< x ≤ 35

2.25 if 35

< x ≤ 45

2.50 if 45

< x ≤ 1.

(c) The cost for a 45

ride is C(45) = $2.25.

Practice Problems

Problem 29.7Make a table of five values of the function f(x) = 2x + 3 and then use thepoints to sketch the graph of f(x).

Problem 29.8Make a table of five values of the function f(x) = 1

2x2 + x and then use the

points to sketch the graph of f(x).

Problem 29.9Make a table of five values of the function f(x) = 3x and then use the pointsto sketch the graph of f(x).

67

Problem 29.10Make a table of five values of the function f(x) = 2 − x3 and then use thepoints to sketch the graph of f(x).

Problem 29.11Which type of function best fits each of the following graphs: linear, quadratic,cubic, exponential, or step?

Problem 29.12Suppose that a function f is given by a table. If the output changes by afixed amount each time the input changes by a constant then the function islinear. Determine whether each of the following functions below are linear.

x 0 2 4 6f(x) 20 40 80 160

x 10 20 30 40g(x) 6 12 18 24

Problem 29.13Show how to solve the equation 2x + 3 = 11 using a calculator.

Problem 29.14In the linear function f(x) = mx + b the parameter m is called the slope.The slope of the line determines whether the line rises, falls, is vertical orhorizontal. Classify the slope of each line as positive, negative, zero, orundefined.

68

Problem 29.15Algebraically, one finds the slope of a line given two points (x1, y1) and (x2, y2)on the line by using the formula

y2 − y1

x2 − x1

.

Would the ratioy1 − y2

x1 − x2

give the same answer? Explain.

Problem 29.16Water is being pumped into a tank. Reading are taken every minutes.

Time(min) 0 3 6 9 12Quarts of water 0 90 180 270 360

(a) Plot the 5 points.(b) What is the slope of the line joining the 5 points?(c) Estimate how much water is in the tank after 1 minute.(d) At what rate is the water being pumped in?

Problem 29.17(a) Graph f(x) = x2 − 2 by plotting the points x = −2,−1, 0, 1, 2.(b) How is this graph related to the graph of y = x2?

Problem 29.18(a) Graph f(x) = x2 + 3 by plotting the points x = −2,−1, 0, 1, 2.(b) How is this graph related to the graph of y = x2?

Problem 29.19(a) Graph f(x) = (x− 1)2 by plotting the points x = −2,−1, 0, 1, 2.(b) How is this graph related to the graph of y = x2?

Problem 29.20(a) Graph f(x) = (x + 2)2 by plotting the points x = −2,−1, 0, 1, 2.(b) How is this graph related to the graph of y = x2?

69

Problem 29.21Graph each equation by plotting points that satisfy the equation.

(a) x− y = 4.(b) y = −2|x− 3|.(c) y = 1

2(x− 1)2.

Problem 29.22Find the x- and y-intercepts of each equation.

(a) 2x + 5y = 12.(b) x = |y| − 4.(c) |x|+ |y| = 4.

70

30 Graphical Representations of Data

Visualization techniques are ways of creating and manipulating graphicalrepresentations of data. We use these representations in order to gain bet-ter insight and understanding of the problem we are studying - pictures canconvey an overall message much better than a list of numbers.In this section we describe some graphical presentations of data.

Line or Dot PlotsLine plots are graphical representations of numerical data. A line plot is anumber line with x’s placed above specific numbers to show their frequency.By the frequency of a number we mean the number of occurrence of thatnumber. Line plots are used to represent one group of data with fewer than50 values.

Example 30.1Suppose thirty people live in an apartment building. These are the followingages:

58 30 37 36 34 49 35 40 47 4739 54 47 48 54 50 35 40 38 4748 34 40 46 49 47 35 48 47 46

Make a line plot of the ages.

Solution.The dot plot is given in Figure 30.1

Figure 30.1

This graph shows all the ages of the people who live in the apartment build-ing. It shows the youngest person is 30, and the oldest is 58. Most people in

71

the building are over 46 years of age. The most common age is 47.

Line plots allow several features of the data to become more obvious. Forexample, outliers, clusters, and gaps are apparent.

• Outliers are data points whose values are significantly larger or smallerthan other values, such as the ages of 30, and 58.• Clusters are isolated groups of points, such as the ages of 46 through 50.• Gaps are large spaces between points, such as 41 and 45.

Practice Problems

Problem 30.1Following are the ages of the 30 students at Washington High School whoparticipated in the city track meet. Draw a dot plot to represent these data.

10 10 11 10 13 8 10 13 14 914 13 10 14 11 9 13 10 11 1211 12 14 13 12 8 13 14 9 14

Problem 30.2The height (in inches) of the players on a professional basketball team are70, 72, 75, 77, 78, 78, 80, 81,81,82, and 83. Make a line plot of the heights.

Problem 30.3Draw a Dot Plot for the following dataset.

50 35 70 55 50 30 4065 50 75 60 45 35 7560 55 55 50 40 55 50

Stem and Leaf PlotsAnother type of graph is the stem-and-leaf plot. It is closely related tothe line plot except that the number line is usually vertical, and digits areused instead of x’s. To illustrate the method, consider the following scoreswhich twenty students got in a history test:

69 84 52 93 61 74 79 65 88 6357 64 67 72 74 55 82 61 68 77

72

We divide each data value into two parts. The left group is called a stemand the remaining group of digits on the right is called a leaf. We displayhorizontal rows of leaves attached to a vertical column of stems. we canconstruct the following table

5 2 7 56 9 1 5 3 4 7 1 87 4 9 2 4 78 4 8 29 3

where the stems are the ten digits of the scores and the leaves are the onedigits.The disadvantage of the stem-and-leaf plots is that data must be groupedaccording to place value. What if one wants to use different groupings? Inthis case histograms are more suited.

If you are comparing two sets of data, you can use a back-to-back stem-and-leaf plot where the leaves of sets are listed on either side of the stemas shown in the table below.

9 6 0 5 78 7 6 4 1 1

8 8 7 6 5 5 3 2 2 2 2 1 2 2 5 6 7 8 8 99 9 6 4 4 3 2 3 1 2 3 4 4 4 5 6 7 8 9

9 6 5 1 4 2 3 5 6 7 8 9 9

Practice Problems

Problem 30.4Given below the scores of a class of 26 fourth graders.

64 82 85 99 96 81 97 80 81 80 84 87 9875 86 88 82 78 81 86 80 50 84 88 83 82

Make a stem-and-leaf display of the scores.

Problem 30.5Each morning, a teacher quizzed his class with 20 geography questions. Theclass marked them together and everyone kept a record of their personal

73

scores. As the year passed, each student tried to improve his or her quizmarks. Every day, Elliot recorded his quiz marks on a stem and leaf plot.This is what his marks looked like plotted out:

0 3 6 51 0 1 4 3 5 6 5 6 8 9 7 92 0 0 0 0

What is his most common score on the geography quizzes? What is hishighest score? His lowest score? Are most of Elliot’s scores in the 10s, 20sor under 10?

Problem 30.6A teacher asked 10 of her students how many books they had read in thelast 12 months. Their answers were as follows:

12, 23, 19, 6, 10, 7, 15, 25, 21, 12

Prepare a stem and leaf plot for these data.

Problem 30.7Make a back-to-back stem and leaf plot for the following test scores:

Class 1:100 96 93 92 92 92 90 9089 89 85 82 79 75 74 7373 73 70 69 68 68 65 6135

Class 2:79 85 56 79 84 64 44 5769 85 65 81 73 51 61 6771 89 69 77 82 75 89 9274 70 75 88 46

Frequency Distributions and HistogramsWhen we deal with large sets of data, a good overall picture and sufficientinformation can be often conveyed by distributing the data into a number ofclasses or class intervals and to determine the number of elements belong-ing to each class, called class frequency. For instance, the following tableshows some test scores from a math class.

74

65 91 85 76 85 87 79 9382 75 100 70 88 78 83 5987 69 89 54 74 89 83 8094 67 77 92 82 70 94 8496 98 46 70 90 96 88 72

It’s hard to get a feel for this data in this format because it is unorganized.To construct a frequency distribution,

• Compute the class width CW = Largest data value−smallest data valueDesirable number of classes

.

• Round CW to the next highest whole number so that the classes coverthe whole data.

Thus, if we want to have 6 class intervals then CW = 100−466

= 9. The lownumber in each class is called the lower class limit, and the high numberis called the upper class limit.With the above information we can construct the following table called fre-quency distribution.

Class Frequency41-50 151-60 261-70 671-80 881-90 1491-100 9

Once frequency distributions are constructed, it is usually advisable to presentthem graphically. The most common form of graphical representation is thehistogram.In a histogram, each of the classes in the frequency distribution is representedby a vertical bar whose height is the class frequency of the interval. The hor-izontal endpoints of each vertical bar correspond to the class endpoints.A histogram of the math scores is given in Figure 30.2

75

Figure 30.2

One advantage to the stem-and-leaf plot over the histogram is that the stem-and-leaf plot displays not only the frequency for each interval, but also dis-plays all of the individual values within that interval.

Practice Problems

Problem 30.8Suppose a sample of 38 female university students was asked their weightsin pounds. This was actually done, with the following results:

130 108 135 120 97 110130 112 123 117 170 124120 133 87 130 160 128110 135 115 127 102 13089 135 87 135 115 110105 130 115 100 125 120120 120

(a) Suppose we want 9 class intervals. Find CW.(b) Construct a frequency distribution.(c) Construct the corresponding histogram.

76

Problem 30.9The table below shows the response times of calls for police service measuredin minutes.

34 10 4 3 9 18 43 14 8 15 19 24 936 5 7 13 17 22 273 6 11 16 21 26 3132 38 40 30 47 53 146 12 18 23 28 333 4 62 24 35 5415 6 13 19 3 44 20 5 4 5 510 25 7 7 42 44

Construct a frequency distribution and the corresponding histogram.

Problem 30.10A nutritionist is interested in knowing the percent of calories from fat whichAmericans intake on a daily basis. To study this, the nutritionist randomlyselects 25 Americans and evaluates the percent of calories from fat consumedin a typical day. The results of the study are as follows

34% 18% 33% 25% 30%42% 40% 33% 39% 40%45% 35% 45% 25% 27%23% 32% 33% 47% 23%27% 32% 30% 28% 36%


Bar GraphsBar Graphs, similar to histograms, are often useful in conveying infor-mation about categorical data where the horizontal scale represents somenonnumerical attribute. In a bar graph, the bars are nonoverlapping rectan-gles of equal width and they are equally spaced. The bars can be vertical orhorizontal. The length of a bar represents the quantity we wish to compare.

Example 30.2The areas of the various continents of the world (in millions of square miles)

77

are as follows:11.7 for Africa; 10.4 for Asia; 1.9 for Europe; 9.4 for NorthAmerica; 3.3 Oceania; 6.9 South America; 7.9 Soviet Union. Draw a barchart representing the above data and where the bars are horizontal.

Solution.The bar graph is shown in Figure 30.3.

Figure 30.3: Areas (in millions of square miles) of thevariuos continents of the world

A double bar graph is similar to a regular bar graph, but gives 2 pieces ofinformation for each item on the vertical axis, rather than just 1. The barchart in Figure 30.4 shows the weight in kilograms of some fruit sold on twodifferent days by a local market. This lets us compare the sales of each fruitover a 2 day period, not just the sales of one fruit compared to another. Wecan see that the sales of star fruit and apples stayed most nearly the same.The sales of oranges increased from day 1 to day 2 by 10 kilograms. Thesame amount of apples and oranges was sold on the second day.

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Figure 30.4

Example 30.3The figures for total population, male and female population of the UK atdecade intervals since 1959 are given below:

Y ear Total UK Resident Population Male Female1959 51, 956, 000 25, 043, 000 26, 913, 0001969 55, 461, 000 26, 908, 000 28, 553, 0001979 56, 240, 000 27, 373, 000 28, 867, 0001989 57, 365, 000 27, 988, 000 29, 377, 0001999 59, 501, 000 29, 299, 000 30, 202, 000

Construct a bar chart representing the data.

Solution.The bar graph is shown in Figure 30.5.

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Figure 30.5

Line GraphsA Line graph ( or time series plot)is particularly appropriate for repre-senting data that vary continuously. A line graph typically shows the trendof a variable over time. To construct a time series plot, we put time on thehorizontal scale and the variable being measured on the vertical scale andthen we connect the points using line segments.

Example 30.4The population (in millions) of the US for the years 1860-1950 is as follows:31.4 in 1860; 39.8 in 1870; 50.2 in 1880; 62.9 in 1890; 76.0 in 1900; 92.0 in1910; 105.7 in 1920; 122.8 in 1930; 131.7 in 1940; and 151.1 in 1950. Make atime plot showing this information.

Solution.The line graph is shown in Figure 30.6.

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Figure 30.6: US population (in millions) from 1860 - 1950

Practice Problems

Problem 30.11Given are several gasoline vehicles and their fuel consumption averages.

Buick 27 mpgBMW 28 mpgHonda Civic 35 mpgGeo 46 mpgNeon 38 mpgLand Rover 16 mpg

(a) Draw a bar graph to represent these data.(b) Which model gets the least miles per gallon? the most?

Problem 30.12The bar chart below shows the weight in kilograms of some fruit sold one

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day by a local market.

How many kg of apples were sold? How many kg of oranges were sold?

Problem 30.13The figures for total population at decade intervals since 1959 are givenbelow:

Y ear Total UK Resident Population1959 51, 956, 0001969 55, 461, 0001979 56, 240, 0001989 57, 365, 0001999 59, 501, 000

Construct a bar chart for this data.

Problem 30.14The following data gives the number of murder victims in the U.S in 1978 clas-sified by the type of weapon used on them. Gun, 11, 910; cutting/stabbing,3, 526; blunt object, 896; strangulation/beating, 1, 422; arson, 255; all others705. Construct a bar chart for this data. Use vertical bars.

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Circle Graphs or Pie ChartsAnother type of graph used to represent data is the circle graph. A circlegraph or pie chart, consists of a circular region partitioned into disjointsections, with each section representing a part or percentage of a whole.To construct a pie chart we first convert the distribution into a percentagedistribution. Then, since a complete circle corresponds to 360 degrees, weobtain the central angles of the various sectors by multiplying the percentagesby 3.6. We illustrate this method in the next example.

Example 30.5A survey of 1000 adults uncovered some interesting housekeeping secrets.When unexpected company comes, where do we hide the mess? The surveyshowed that 68% of the respondents toss their mess in the closet, 23% shovethings under the bed, 6% put things in the bath tub, and 3% put the messin the freezer. Make a circle graph to display this information.

Solution.We first find the central angle corresponding to each case:

in closet 68× 3.6 = 244.8under bed 23× 3.6 = 82.8in bathtub 6× 3.6 = 21.6in freezer 3× 3.6 = 10.8

Note that244.8 + 82.8 + 21.6 + 10.8 = 360.

The pie chart is given in Figure 30.7.

Figure 30.7: Where to hide the mess?

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Practice Problems

Problem 30.15The table below shows the ingredients used to make a sausage and mushroompizza.

Ingredient %Sausage 7.5Cheese 25Crust 50

TomatoSause 12.5Mushroom 5

Plot a pie chart for this data.

Problem 30.16A newly qualified teacher was given the following information about theethnic origins of the pupils in a class.

Ethnic origin No. of pupilsWhite 12Indian 7

BlackAfrican 2Pakistani 3

Bangladeshi 6TOTAL 30

Plot a pie chart representing the data.

Problem 30.17The following table represent a survey of people’s favorite ice cream flavor

Flovor Number of peopleV anilla 21.0%

Chocolate 33.0%Strawberry 12.0%Raspberry 4.0%

Peach 7.0%Neopolitan 17.0%

Other 6.0%

Plot a pie chart to represent this data.

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Problem 30.18In the United States, approximately 45% of the population has blood typeO; 40% type A; 11% type B; and 4% type AB. Illustrate this distribution ofblood types with a pie chart.

PictographsOne type of graph seen in newspapers and magazines is a pictograph. Ina pictograph, a symbol or icon is used to represent a quantity of items. Apictograph needs a title to describe what is being presented and how the dataare classified as well as the time period and the source of the data. Exampleof a pictograph is given in Figure 30.8.

Figure 30.8

A disadvantage of a pictograph is that it is hard to quantify partial icons.

Practice Problems

Problem 30.19

Make a pictograph to represent the data in the following table. Use torepresent 10 glasses of lemonade.

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Day FrequencyMonday 15Tuesday 20Wednesday 30Thursday 5Friday 10

Problem 30.20The following pictograph shows the approximate number of people who speakthe six common languages on earth.(a) About how many people speak Spanish?(b) About how many people speak English?(c) About how many more people speak Mandarin than Arabic?

Problem 30.21Twenty people were surveyed about their favorite pets and the result is shownin the table below.

Pet FrequencyDog 2Cat 5Hamster 3

Make a pictograph for the following table of data. Let stand for 2 votes.

ScatterplotsA relationship between two sets of data is sometimes determined by usinga scatterplot. Let’s consider the question of whether studying longer for atest will lead to better scores. A collection of data is given below

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Study Hours 3 5 2 6 7 1 2 7 1 7Score 80 90 75 80 90 50 65 85 40 100

Based on these data, a scatterplot has been prepared and is given in Figure30.9.(Remember when making a scatterplot, do NOT connect the dots.)

Figure 30.9

The data displayed on the graph resembles a line rising from left to right.Since the slope of the line is positive, there is a positive correlation be-tween the two sets of data. This means that according to this set of data,the longer I study, the better grade I will get on my exam score.If the slope of the line had been negative (falling from left to right), a neg-ative correlation would exist. Under a negative correlation, the longer Istudy, the worse grade I would get on my exam.If the plot on the graph is scattered in such a way that it does not approx-imate a line (it does not appear to rise or fall), there is no correlationbetween the sets of data. No correlation means that the data just doesn’tshow if studying longer has any affect on my exam score.

Practice Problems

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Problem 30.22Coach Lewis kept track of the basketball team’s jumping records for a 10-yearperiod, as follows:

Year ’93 ’94 ’95 ’96 ’97 ’98 ’99 ’00 ’01 ’02Record(nearest in) 65 67 67 68 70 74 77 78 80 81

(a) Draw a scatterplot for the data.(b) What kind of corrolation is there for these data?

Problem 30.23The gas tank of a National Motors Titan holds 20 gallons of gas. The fol-lowing data are collected during a week.

Fuel in tank (gal.) 20 18 16 14 12 10Dist. traveled (mi) 0 75 157 229 306 379


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31 Misleading Graphs and Statistics

It is a well known fact that statistics can be misleading. They are often usedto prove a point, and can easily be twisted in favour of that point! The pur-pose of this section is to learn how to recognize common statisitcal deceptionso that to avoid being mislead.

Bad SamplingWhen you use a sample to represent a larger group, you must make sure thatthe people in the sample are fairly representative of the larger group.

Example 31.1Decide whether a mall is a good place to find a sample for a survey aboutthe amount of allowance received by people ages 10 to 15.

Solution.The mall is probably not a representative place to find a fair sample of peo-ple in this age range. Taking a sample at the mall might not represent fairlythose people who receive a small allowance, or none

Misleading GraphsGood graphs are extremely powerful tools for displaying large quantities ofcomplex data; they help turn the realms of information available today intoknowledge. But, unfortunately, some graphs deceive or mislead. This mayhappen because the designer chooses to give readers the impression of betterperformance or results than is actually the situation. In other cases, the per-son who prepares the graph may want to be accurate and honest, but maymislead the reader by a poor choice of a graph form or poor graph construc-tion.The following things are important to consider when looking at a graph:

1. Title2. Labels on both axes of a line or bar chart and on all sections of a pie chart3. Source of the data4. Key to a pictograph5. Uniform size of a symbol in a pictograph6. Scale: Does it start with zero? If not, is there a break shown7. Scale: Are the numbers equally spaced?

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Scaling and Axis ManipulationA graph can be altered by changing the scale of the graph. For example,data in the two graphs of Figure 31.1 are identical, but scaling of the Y-axischanges the impression of the magnitude of differences.

Figure 31.1

Example 31.2Why does the bar chart below misleading? How should the information berepresented?

Solution.The bar chart indicates that house prices have tripled in one year. The scaleof vertical must start at 0 and that’s not the case. A less misleading graphwould look like the one in Figure 31.2. This gives a much more accuratepicture of what has happened.

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Figure 31.2

Example 31.3What is wrong with the information represented on this graph?

Solution.Although the vertical scale starts at 0, it does not go up in even steps. Thisdistorts the graph, and makes it look as though the biggest jump is between1 and 2 rather than 3 and 4. Also, there are no labels on the axes so we haveno idea what this graph represents!

Three Dimensional Effects

Example 31.4What is wrong with this 3D bar chart?

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Solution.This 3D bar chart might look very attractive, but it is also very misleading.There is no scale on the vertical axis, and because of the perspective it looksas though the sales for 1995 were far greater than those for any other year.In fact they were identical to those for 1997. It would be much better todraw a 2D bar chart like the one shown in Figure 31.3 with the appropriatelabelling on each axis:

Figure 31.3

Deceptive Pictographs

Example 31.5What is wrong with this pictogram showing the number of people who owndifferent types of pets?

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Solution.On this pictogram there isn’t a category for those people who do not own apet. The pictures are different sizes and it appears that more people own ahorse than any other animal.An improvement would be to redraw the pictogram with each of the animalsthe same size and aligned with one another as shown in Figure 31.4.

Figure 31.4

Example 31.6A survey was conducted to determine what food would be served at the

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French club party. Explain how the graph misrepresents the data.

Solution.The percents on the circle graph do not sum to 100.

Example 31.7The number of graduates from a community college for the years 1999 through2003 is given in the following table:

Year 1999 2000 2001 2002 2003# of Graduates 140 180 200 210 160

The figure below shows the line graphs of the same data but with differentscales. Comment on that.

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Solution.The two graphs do not convey the same message. In Figure (b) the spac-ing of the years on the horizontal axis is more spread out and that for thenumbers on the vertical axis is more condensed than Figure (a). A collegeadministrator might use a graph like Figure (b) to convince people that thecollege was not in serious enrollment trouble.

Practice Problems

Problem 31.1Jenny averaged 70 on her quizzes during the first part of the quarter and 80on her quizzes during the second part of the quarter. When she found outthat her final average for the quarter was not 75, she went to argue with herteacher. Give a possible explanation for Jenny’s misunderstanding.

Problem 31.2Suppose the following circle graphs are used to illustrate the fact that thenumber of elementary teaching majors at teachers’ colleges has doubled be-tween 1993 and 2003, while the percent of male elementary teaching majorshas stayed the same. What is misleading about the way the graphs are con-structed?

Problem 31.3What is wrong with the following line graph?

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Problem 31.4Doug’s Dog Food Company wanted to impress the public with the magnitudeof the company’s growth. Sales of Doug’s Dog Food had doubled from 2002to 2003, so the company displayed the following graph, in which the radiusof the base and the height of the 2003 can are double those of the 2002 can.What does the graph really show with respect to the growth of the company?

Problem 31.5What’s wrong with the following graph?

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Problem 31.6Refer to the following pictograph:

Ms McNulty claims that on the basis of this information, we can concludethat men are worse drivers than women. Discuss whether you can reachthat conclusion from the pictograph or you need more information. If moreinformation is needed, what would you like to know?

Problem 31.7Larry and Marc took the same courses last quarter. Each bet that he wouldreceive the better grades. Their courses and grades are as follows:

Course Larry’s Grades Marc’s GradesMath(4 credits) A CChemistry(4 credits) A CEnglish(3 credits) B BPsychology(3 credits) C ATennis(1 credit) C A

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Marc claimed that the results constituted a tie, since both received 2 A’s,1 B, and 2 C’s. Larry said that he won the bet because he had the higherGPA for the quarter. Who is correct?(Allow 4 points for A, 3 points for B,2 points for C, 1 point for D, and 0 point for F.)

Problem 31.8Oil prices went up 20% one year and 30% the next. Is it true that over thetwo years, prices went up 50%?

Problem 31.9True or false? My rent went down 10% last year and then rose 20% this year.Over the two years my rent went up by 10%.

Problem 31.10Which graph could be used to indicate a greater decrease in the price ofgasoline? Explain.

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32 Measures of Central Tendency and Dis-

persion

In this section we discuss two important aspects of data which are its centerand its spread. The mean, median, and the mode are measures of centraltendency that describe where data are centered. The range, variance, andstandard deviation are measures of dispersion that describe the spread ofdata.

Measures of Central Tendency: Mode, Median, Mean

ModeThe first measure of the center is the mode. It is defined as the value whichoccurs with the highest frequency in the data. Thus, it is used when oneis interested in the most common value in a distribution. A mode may notexist and usually this happens when no data value occurs more frequentlythan all the others. Also a mode may not be unique.

Example 32.1The final grades of a class of six graduate students were A, C,B,C, A, B.What is the mode?

Solution.Since the grades occur at the same frequency then there is no mode for thisdata.

Example 32.2A sample of the records of motor vehicle bureau shows that 18 drivers in acertain age group received 3, 2, 0, 0, 2, 3, 3, 1, 0, 1, 0, 3, 4, 0, 3, 2, 3, 0 traffic tick-ets during the last three year. Find the mode?

Solution.The mode consists of 0 and 3 since they occur six times on the list.

MedianAnother measure of the center is the median. The median usually is foundwhen we have an ordered distribution. It is computed as follows. We arrange

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the numerical data from smallest to largest. If n denotes the size of the setof data then the median can be found by using the median rank

MR =n + 1

2.

If MR is a whole number then the median is the value in that position. IfMR ends in .5, we take the sum of the adjacent positions and divide by 2.Unlike the mode, the median always exists and is unique. But it may or maynot be one of the given data values. Note that extreme values (smallest orlargest) do not affect the median.

Example 32.3Among groups of 40 students interviewed at each of 10 different colleges,18, 13, 15, 12, 8, 3, 7, 14, 16, 3 said that they jog regularly. Find the median.

Solution.First, arrange the numbers from smallest to largest to obtain

3 3 7 8 12 13 14 15 16 18

Next, compute the median rank MR = 10+12

= 5.5. Hence, the median is12+13

2= 12.5

Example 32.4Nine corporations reported that in 1982 they made cash donations to 9, 16, 11,10, 13, 12, 6, 9, and 12 colleges. Find the median number.

Solution.Arranging the numbers from smallest to largest to obtain

6 9 9 10 11 12 12 13 16

The median rank is MR = 9+12

= 5. The median is 11.

Arithmetic MeanAnother most widely used measure of the center is the arithmetic meanor simply mean. The mean of a set of N numbers x1, x2, ..., xN , denoted byx, is defined as

x =x1 + x2 + ... + xN

N.

Unlike the median, the mean can be affected by extreme values since it usesthe exact value of each data.

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Example 32.5If nine school juniors averaged 41 on the verbal portion of the PSAT test, atmost how many of them can have scored 65 or more?

Solution.We have that x = 41 and N = 9 so that x1 + x2 + · · · + x9 = 41× 9 = 369.Since 6×65 = 390 > 369 and 5×65 = 325 then at most 5 students can scoremore than 65.

Example 32.6If the numbers x1, x2, ..., xN occur with frequencies m1, m2, ...,mN respec-tively then what is the mean in this case?

Solution.The mean is given by

x =m1x1 + m2x2 + ... + mNxN

m1 + m2 + · · ·+ mN

.

The figure below gives the relationships among the measures of central ten-dency.

Practice Problems

Problem 32.1Find (a) the mean, (b) median, and (c) the mode for the following collectionof data:

60 60 70 95 95 100

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Problem 32.2Suppose a company employs 20 people. The president of the company earns$200,000, the vice president earns $75,000, and 18 employees earn $10,000each. Is the mean the best number to choose to represent the ”average”salary of the company?

Problem 32.3Suppose nine students make the following scores on a test:

30, 35, 40, 40, 92, 92, 93, 98, 99

Is the median the best ”average” to represent the set of scores?

Problem 32.4Is the mode an appropriate ”average” for the following test scores?

40, 42, 50, 62, 63, 65, 98, 98

Problem 32.5The 20 meetings of a square dance club were attended by 26, 25, 28, 23, 25,24, 24, 23, 26, 26, 28, 26, 24, 32, 25, 27, 24, 23, 24, and 22 of its members.Find the mode, median, and mean.

Problem 32.6If the mean annual salary paid to the top of three executives of a firm is$96, 000, can one of them receive an annual salary of $300, 000?

Problem 32.7An instructor counts the final examination in a course four times as much asof the four one-hour examinations. What is the average grade of a studentwho received grades of 74, 80, 61, and 77 in the four one- hour examinationsand 83 in the final examination?

Problem 32.8In 1980 a college paid its 52 instructors a mean salary of $13, 200, its 96assistant professors a mean salary of $15, 800, its 67 associate professors amean salary of $18, 900, and its 35 full professors a mean salary of $23, 500.What was the mean salary paid to all the teaching staff of this college?

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Measures of Dispersion: Range, Variance, and Standard DeviationWhile mean and median tell you about the center of your observations, theysay nothing about how data are scattered. variability or dispersion mea-sures the extent to which data are spread out.The measures of variability for data that we look at are: the range, the in-terquartile range, the variance, and the standard deviation.

The RangeTo measure the variability between extreme values (i.e. smallest and largestvalues) one uses the range. The range is the difference between the largestand smallest values of a distribution.

Example 32.7Find the range of each of the following samples:

Sample 1: 6,18,18,18,18,18,18,18,18,18.Sample 2: 6,6,6,6,6,6,18,18,18,18,18.Sample 3: 6,7,9,11,12,14,15,16,17,18.

Solution.Sample 1: 18 - 6 = 12Sample 2: 18 - 6 = 12Sample 3: 18 - 6 = 12.

As you can see from this example, each sample has a range 18 − 6 = 12but the spread out of values is quite different in each case. In Sample 1, thespread is uniform whereas it is not in Sample 3. This is a disadvantage ofthis kind of measure. The range tells us nothing about the dispersion of thevalues between the extreme (smallest and largest) values. A better under-standing is obtained by determining quartiles.

Quartiles and PercentilesRecall that if a set of data is arranged from smallest to largest, the middlevalue Q2 (or the arithmetic mean of the two middle values) that divides theset of observations into two equal parts I1 and I2 is called the median ofthe distribution. That is, 50 percent of the observations are larger than themedian and 50 percent are smaller.

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Now, the median of I1 is denoted by Q1 (called the lower quartile) and thatof I2 by Q3 (called the upper quartile). Thus, Q1, Q2 and Q3 divide theset of data into four equal parts. We call Q1, Q2 and Q3 the three quartilesof the distribution.In a similar manner, one could construct other measures of variation by con-sidering percentiles rather than quartiles. For a whole number P , where1 ≤ P ≤ 99, the Pth percentile of a distribution is a value such that P% ofthe data fall at or below it. Thus, there are 99 percentiles. The percentileslocations are found using the formula

LP = (n + 1) · P

100.

Thus, the median Q2 is the 50th percentile so that P = 50 and L50 = n+12

which is the median rank. Note that Q1 is the 25th percentile and Q3 is the75th percentile.An example will help to explain further.

Example 32.8Listed below are the commissions earned, in dollars, last month by a sampleof 15 brokers at Salomon Smith Barneys office.

2038 1758 1721 1637 20972047 2205 1787 2287 19402311 2054 2406 1471 1460

(a) Rank the data from smallest to largest.(b) Find the median rank and then the median.(c) Find the quartiles Q1 and Q3.

Solution.(a) Arranging the data from smallest to largest we find

1460 1471 1637 1721 17581787 1940 2038 2047 20542097 2205 2287 2311 2406

(b) The median Q2 is the 50th percentile so that P = 50 and L50 = 15+12

= 8.Thus, Q2 = 2038.

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(c) Q1 is the 25th percentile so that L25 = (15+1)· 25100

= 4 and so Q1 = 1721.Similarly, Q3 is the 75th percentile so that P = 75 and L75 = (15+1) · 75

100=

12. Thus, Q3 = 2205.

Percentiles are extensively used in such fields as educational testing. Un-doubtedly some of you have had the experience of being told at what per-centile you rated on a scholastic aptitude test.

Example 32.9You took the English achievement test to obtain college credit in freshmanEnglish by examination. If your score was in the 89th percentile, what doesthis mean?

Solution.This means 89% of the scores were at or below yours.

Remark 32.1In Example 32.8, LP was found to be a whole number. That is not alwaysthe case. For example, if n = 20 observations then L25 = (20+1) · 25

100= 5.25.

In this case, to find Q1 we locate the fifth value in the ordered array and thenmove .25 of the distance between the fifth and sixth values and report thatas the first quartile. Like the median, the quartile does not need to be oneof the actual values in the data set.To explain further, suppose a data set contained the six values: 91, 75, 61,101, 43, and 104. We want to locate the first quartile. We order the valuesfrom smallest to largest: 43, 61, 75, 91, 101, and 104. The first quartile islocated at

L25 = (6 + 1) · 25

100= 1.75.

The position formula tells us that the first quartile is located between thefirst and the second value and that it is .75 of the distance between the firstand the second values. The first value is 43 and the second is 61. So thedistance between these two values is 18. To locate the first quartile, we needto move .75 of the distance between the first and second values, so .75(18) =13.5. To complete the procedure, we add 13.5 to the first value and reportthat the first quartile is 56.5.

Box-and-Whisker PlotA box-and-whisker plot is a graphical display, based on quartiles, that

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helps us picture a set of data. The steps for making such a box are asfollows:

• Draw a vertical (or horizontal) scale to include the lowest and highestdata values.

• To the right (or upper for the horizontal scale) of the scale draw a boxfrom Q1 to Q3.

• Include a solid line through the box at the median level.

• Draw solid lines, called whiskers, from Q1 to the lowest value andfrom Q3 to the largest value.

An example will help to explain.

Example 32.10Alexanders Pizza offers free delivery of its pizza within 15 km. Alex, theowner, wants some information on the time it takes for delivery. How longdoes a typical delivery take? Within what range of times will most deliveriesbe completed? For a sample of 20 deliveries, he determined the followinginformation:

Minimum value = 13 minutesQ1 = 15 minutesMedian = 18 minutesQ3 = 22 minutesMaximum value = 30 minutes

Develop a box plot for the delivery times. What conclusions can you makeabout the delivery times?

Solution.The box plot is shown in Figure 32.1.

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Figure 32.1

The box plot shows that the middle 50 percent of the deliveries take between15 minutes and 22 minutes. The distance between the ends of the box, 7minutes, is the interquartile range. The interquartile range is the distancebetween the first and the third quartile. It tells us the spread of the middlehalf of the data.The box plot also reveals that the distribution of delivery times is positivelyskewed since the median is not in the center of the box and the distance fromthe first quartile to the median is smaller than the distance from the medianto the third quartile.

Practice Problems

Problem 32.9The following table gives the average costs of a single-lens reflex camera:

800 650 300 430 560 470 640 830400 280 800 410 360 600 310 370

(a) Rank the data from smallest to largest.(b) Find the quartiles Q1, Q2, and Q3.(c) Make a box-and-whisker plot.

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Problem 32.10Mr. Eyha took a general aptitude test and scored in the 82nd percentile foraptitude in accounting. What percentage of the scores were at or below hisscore? What percentage were above?

Problem 32.11At Center Hospital there is a concern about the high turnover of nurses. Asurvey was done to determine how long (in months) nurses had been in theircurrent positions. The responses of 20 nurses were:

23 2 5 14 25 36 27 42 12 87 23 29 26 28 11 20 31 8 36

(a) Rank the data.(b) Make a box-and-whisker plot of the data.(c) What are your conclusions from the plot?

Variance and Standard DeviationWe have seen that a remedy for the deficiency of the range is the use ofbox and whisker plot. However, an even better measure of variability is thestandard deviation. Unlike the range, the variance combines all the values ina data set to produce a measure of spread. The variance and the standarddeviation are both measures of the spread of the distribution about the mean.If µ is the population mean of set of data then the quantity (x− µ) is calledthe deviation from the mean. The variance of a data set is the arithmeticaverage of squared deviation from the mean. It is denoted by s2 and is givenby the formula

s2 =

∑ni=1(xi − µ)2

n.

Note that the variance is nonnegative, and it is zero only if all observationsare the same.The standard deviation is the square root of the variance:

σ =

√∑ni=1(xi − µ)2

n.

The variance is the nicer of the two measures of spread from a mathematicalpoint of view, but as you can see from the algebraic formula, the physicalunit of the variance is the square of the physical unit of the data. For exam-ple, if our variable represents the weight of a person in pounds, the variance

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measures spread about the mean in squared pounds. On the other hand,standard deviation measures spread in the same physical unit as the originaldata, but because of the square root, is not as nice mathematically. Bothmeasures of spread are useful.

A step by step approach to finding the standard deviation is:

1. Calculate the mean.2. Subtract the mean from each observation.3. Square each result.4. Add these squares.5. Divide this sum by the number of observations.6. Take the positive square root.

The variance and standard deviation introduced above are for a population.We define the sample variance by the formula

s2 =

∑ni=1(xi − x)2

n− 1

and the sample standard deviation by the formula

s =

√∑ni=1(xi − x)2

n− 1.

The reason that for s we use n−1 instead of n is because usually a sample doesnot contain extreme values and we want s to be an estimate of σ thereforeby using n− 1 we make s a little larger than if we divide by n.

Remark 32.2Note that if the standard deviation is large than the data are more spreadout whereas when the standard deviation is small then the data are moreconcentrated near the mean.

Example 32.11The owner of the Ches Tahoe restaurant is interested in how much peoplespend at the restaurant. He examines 10 randomly selected receipts forparties of four and writes down the following data.

44 50 38 96 42 47 40 39 46 50

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(a) Find the arithmetic mean.(b) Find the variance and the standard deviation.

Solution.(a) The arithmetic mean is the sum of the above values divided by 10,i.e.,x = 49.2Below is the table for getting the standard deviation:

x x− 49.2 (x− 49.2)2

44 -5.2 27.0450 0.8 0.6438 11.2 125.4496 46.8 2190.2442 -7.2 51.8447 -2.2 4.8440 -9.2 84.6439 -10.2 104.0446 -3.2 10.2450 0.8 0.64Total 2600.4

Hence the variance is s2 = 2600.49

≈ 289 and the standard deviation is

s =√

289 = 17. What this means is that most of the patrons probablyspend between 49.20 - 17 =$32.20 and 49.20+17=$66.20.

Practice Problems

Problem 32.12The following are the wind velocities reported at 6 P.M.on six consecutivedays: 13,8,15,11,3 and 10. Find the range, sample mean, sample variance,and sample standard deviation.

Problem 32.13An airline’s records show that the flights between two cities arrive on theaverage 4.6 minutes late with a standard deviation of 1.4 minutes. At leastwhat percentage of its flights between these two cities arrive anywhere be-tween 1.8 minutes late and 7.4 minutes late?

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Problem 32.14One patient’s blood pressure, measured daily over several weeks, averaged182 with a standard deviation of 5.3, while that of another patient averaged124 with a standard deviation of 9.4. Which patient’s blood pressure isrelatively more variable?

Problem 32.15By sampling different landscapes in a national park over a 2-year period, thenumber of deer per square kilometer was determined. The results were (deerper square kilometer)

30 20 5 29 58 720 18 4 29 22 9

Compute the range, sample mean, sample variance, and sample standarddeviation.

Problem 32.16A researcher wants to find the number of pets per household. The researcherconducts a survey of 35 households. Find the sample variance and standarddeviation.

0 2 3 1 01 2 3 1 01 2 1 1 03 2 1 1 14 1 2 2 43 2 1 2 32 3 4 0 2

Problem 32.17Suppose two machines produce nails which are on average 10 inches long. Asample of 11 nails is selected from each machine.

Machine A: 6, 8, 8, 10, 10, 10, 10, 10, 12, 12, 14.Machine B: 6, 6, 6, 8, 8, 10, 12, 12, 14, 14, 14.

Which machine is better than the other?

Problem 32.18Find the missing age in the following set of four student ages.

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Student Age Deviation from the MeanA 19 -4B 20 -3C ? 1D 29 6

Problem 32.19The maximum heart rates achieved while performing a particular aerobicexercise routine are measured (in beats per minute) for 9 randomly selectedindividuals.

145 155 130 185 170 165 150 160 125

(a) Calculate the range of the time until failure.(b) Calculate the sample variance of the time until failure.(c) Calculate the sample standard variation of the time until failure.

Problem 32.20The following data gives the number of home runs that Babe Ruth hit ineach of his 15 years with the New York Yankees baseball team from 1920 to1934:

54 59 35 41 46 25 47 60 54 46 49 46 41 34 22.

The following are the number of home runs that Roger Maris hit in each ofthe ten years he played in the major leagues from 1957 on:

8 13 14 16 23 26 28 33 39 61

Calculate the mean and standard deviation for each player’s data and com-ment on the consistency of performance of each player.

Problem 32.21An office of Price Waterhouse Coopers LLP hired five accounting traineesthis year. Their monthly starting salaries were: $2536; $2173; $2448; $2121;and $2622.

(a) Compute the population mean.(b) Compute the population variance.(c) Compute the population standard deviation.

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Normal DistributionTo better understand how standard deviations are used as measures of dis-persion, we next consider normal distributions. The graph of a normal dis-tribution is called a normal curve or a bell-shaped curve. The curve hasthe following properties:

1. The curve is bell-shaped with the highest point over the mean µ.

2. It is symmetrical about the line through µ.

3. The curve approaches the horizontal axis but never touches or crossesit.

4. The points where the curve changes concavity occur at µ+σ and µ−σ.

5. The total area under the curve is assumed to be 1.(0.5 to the left ofthe mean and 0.5 to the right).

The data for a normal distribution are spread according to the following rules(See Figure 32.2):

• About 68 percent of the data values will lie within one standard devi-ation of the mean.

• About 95 percent of the data values will lie within two standard devi-ations of the mean.

• About 99.7 percent of the data values will lie within three standarddeviations of the mean.

This result is sometimes referred to as the emperical rule, because thegiven percentages are observed in practice.

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Figure 32.2

Example 32.12When a standardized test was scored, there was a mean of 500 and a stan-dard deviation of 100. Suppose that 10,000 students took the test and theirscores had a bell-shaped distribution.

(a) How many scored between 400 and 600?(b) How many scored between 300 and 700?(c) How many scored between 200 and 800?

Solution.(a) Since one standard deviation on either side of the mean is from 400 to600, about 68% of the scores fall in this interval. Thus, 0.68×10, 000 = 6800students scored between 400 and 600.(b) About 95% of 10,000 or 9500 students scored between 300 and 700.(c) About 99.7% of 10,000 or 9970 students scored between 200 and 800

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By the discussion above we get information of the percentage of data withina certain number of standard deviations. However, if we want to find thelocation of a data value x from the mean we can use the so-called z-scoregiven by the formula

z =x− µ

σ

Now, if the z-score is given then the raw data can be found by solving theequation z = x−µ

σfor x to obtain

x = µ + zσ.

Note that from this formula we see that the value of z tells us how manystandard deviation the corresponding value of x lies above (if z > 0) or below(if z < 0) the mean of its distribution.

Example 32.13Scores on intelligence tests (IQs) are normally distributed in children. IQsfrom the Wechsler intelligence tests are known to have means of 100 andstandard deviations of 15. In almost all the states in the United States(Pennsylvania and Nebraska are exceptions) children can be labeled as men-tally retarded if their IQ falls to 70 points or below. What is the maximumz score one could obtain on an intelligence test and still be considered to bementally retarded?

Solution.Applying the z-score formula we find z = 70−100

15= −2.

Example 32.14In a certain city the mean price of a quart of milk is 63 cents and the standarddeviation is 8 cents. The average price of a package of bacon is $1.80 and thestandard deviation is 15 cents. If we pay $0.89 for a quart of milk and $2.19for a package of bacon at a 24-hour convenience store, which is relativelymore expensive?

Solution.To answer this, we compute z-scores for each:

zMilk =0.89− 0.63

0.08= 3.25

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and

zBacon =2.19− 1.80

0.15= 2.60.

Our z-scores show us that we are overpaying quite a bit more for the milkthan we are for the bacon.

Example 32.15Graduate Record Examination (GRE) scores have means equal to 500 andstandard deviations of 100. If a person receives a z-score on the GRE of 1.45,what would their raw score be?

Solution.Using the formula x = µ + zσ we find x = 500 + 1.45(100) = 645.

Practice Problems

Problem 32.22On a final examination in Statistics, the mean was 72 and the standarddeviation was 15. Assuming normal distribution, determine the z-score ofstudents receiving the grades (a) 60, (b) 93, and (c) 72.

Problem 32.23Referring to the previous exercise, find the grades corresponding to the z-score z = 1.6.

Problem 32.24If z1 = 0.8, z2 = −0.4 and the corresponding x-values are x1 = 88 andx2 = 64 then find the mean and the standard deviation, assuming we have anormal distribution.

Problem 32.25A student has computed that it takes an average (mean) of 17 minutes witha standard deviation of 3 minutes to drive from home, park the car, and walkto an early morning class. Assuming normal distribution,

(a) One day it took the student 21 minutes to get to class. How manystandard deviations from the average is that?(b) Another day it took only 12 minutes for the student to get to class. Whatis this measurement in standard units?(c) Another day it took him 17 minutes to be in class. What is the z-score?

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Problem 32.26Mr. Eyha’s z-score on a college exam is 1.3. If the x-scores have a mean of480 and a standard deviation of 70 points, what is his x-score?

Problem 32.27(a) If µ = 80, σ = 10, what is the z-score for a person with a score of 92?(b) If µ = 65, σ = 12, what is the raw score for a z-score of -1.5?

Problem 32.28Sketch a normal curve. Mark the axis corresponding to the parameter µ andthe axis corresponding to µ + σ and µ− σ.

Problem 32.29For the population of Canadian high school students, suppose that the num-ber of hours of TV watched per week is normally distributed with a mean of20 hours and a standard deviation of 4 hours. Approximately, what percent-age of high school students watch

(a) between 16 and 24 hours per week?(b) between 12 and 28 hours per week?(c) between 8 and 32 hours per week?

Problem 32.30The length of human pregnancies from conception to birth varies accord-ing to a distribution that is approximately normal with mean 266 days andstandard deviation 16 days. Use the emperical rule to answer the followingquestions.

(a) Between what values do the lengths of the middle 95% of all pregnanciesfall?(b) How short are the shortest 2.5% of all pregnancies?

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33 Probability: Some Basic Terms

In this and the coming sections we discuss the fundamental concepts of prob-ability at a level at which no previous exposure to the topic is assumed.Probability has been used in many applications ranging from medicine tobusiness and so the study of probability is considered an essential compo-nent of any mathematics curriculum.So what is probability? Before answering this question we start with somebasic definitions.

An experiment is any situation whose outcome cannot be predicted withcertainty. Examples of an experiment include rolling a die, flipping a coin,and choosing a card from a deck of playing cards.By an outcome or simple event we mean any result of the experiment.For example, the experiment of rolling a die yields six outcomes, namely, theoutcomes 1,2,3,4,5, and 6.The sample space S of an experiment is the set of all possible outcomesfor the experiment. For example, if you roll a die one time then the exper-iment is the roll of the die. A sample space for this experiment could beS = {1, 2, 3, 4, 5, 6} where each digit represents a face of the die.An event is a subset of the sample space. For example, the event of rollingan odd number with a die consists of three simple events {1, 3, 5}.

Example 33.1Consider the random experiment of tossing a coin three times.

(a) Find the sample space of this experiment.(b) Find the outcomes of the event of obtaining more than one head.

Solution.(a) The sample space is composed of eight simple events:

S = {TTT, TTH, THT, THH,HTT, HTH,HHT, HHH}.

(b) The event of obtaining more than one head is the set {THH,HTH, HHT, HHH}.

Example 33.2Consider the random experiment of rolling a die.

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(a) Find the sample space of this experiment.(b) Find the event of rolling the die an even number.

Solution.(a) The sample space is composed of six simple events:

S = {1, 2, 3, 4, 5, 6}.

(b) The event of rolling the die an even number is the set {2, 4, 6}.

Example 33.3An experiment consists of the following two stages: (1) first a fair die isrolled and the number of dots recorded, (2) if the number of dots appearingis even, then a fair coin is tossed and its face recorded, and if the number ofdots appearing is odd, then the die is tossed again, and the number of dotsrecorded. Find the sample space of this experiment.

Solution.The sample space for this experiment is the set of 24 ordered pairs

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, H), (2, T ),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, H), (4, T ),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, H), (6, T )} .

Probability is the measure of occurrence of an event. If the event is impos-sible to occur then its probability is 0. If the occurrence is certain then theprobability is 1. The closer to 1 the probability is, the more likely the eventis. The probability of occurrence of an event E (called its success) will bedenoted by P (E). If an event has no outcomes, that is as a subset of S ifE = ∅ then P (∅) = 0. On the other hand, if E = S then P (S) = 1.

Example 33.4A hand of 5 cards is dealt from a deck. Let E be the event that the handcontains 5 aces. List the elements of E.

Solution.Recall that a standard deck of 52 playing cards can be described as follows:

hearts (red) Ace 2 3 4 5 6 7 8 9 10 Jack Queen Kingclubs (black) Ace 2 3 4 5 6 7 8 9 10 Jack Queen Kingdiamonds (red) Ace 2 3 4 5 6 7 8 9 10 Jack Queen Kingspades (black) Ace 2 3 4 5 6 7 8 9 10 Jack Queen King

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Cards labeled Jack, Queen, or King are called face cards.Since there are only 4 aces in the deck, event E cannot occur. Hence E is animpossible event and E = ∅ so that P (E) = 0

When the outcome of an experiment is just as likely as another, as in theexample of tossing a coin, the outcomes are said to be equally likely. Var-ious probability concepts exist nowadays. The classical probability conceptapplies only when all possible outcomes are equally likely, in which case weuse the formula

P (E) =number of outcomes favorable to event

total number of outcomes=

n(E)

n(S),

where n(E) denotes the number of elements in E.Since for any event E we have ∅ ⊆ E ⊆ S then 0 ≤ n(E) ≤ n(S) so that

0 ≤ n(E)n(S)

≤ 1. It follows that 0 ≤ P (E) ≤ 1.

Example 33.5Which of the following numbers cannot be the probability of some event?(a) 0.71 (b)−0.5 (c) 150% (d) 4

3.

Solution.Only (a) can represent the probability of an event. The reason that −0.5 isnot because it is a negative number. As for 150% this is a number greaterthan 1 and so cannot be a probability of an event. The same is true for 4

3.

Example 33.6What is the probability of drawing an ace from a well-shuffled deck of 52playing cards?

Solution.Since there are four aces in a deck of 52 playing cards then the probabiltiyof getting an ace is 4

52= 1

13.

Example 33.7What is the probability of rolling a 3 or a 4 with a fair die?

Solution.Since the event of having a 3 or a 4 has two simple events {3, 4} then the

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probability of rolling a 3 or a 4 is 26

= 13.

It is important to keep in mind that the above definition of probability ap-plies only to a sample space that has equally likely outcomes. Applying thedefinition to a space with outcomes that are not equally likely leads to incor-rect conclusions. For example, the sample space for spinning the spinner inFigure 33.1 is given by S={Red,Blue}, but the outcome Blue is more likely tooccur than is the outcome Red. Indeed, P (Blue) = 3

4whereas P (Red) = 1

4.

Figure 33.1

A widely used probability concept is the experimental probability whichuses the relative frequency of an event and is given by the formula:

P (E) = Relative frequency =f

n,

where f is the frequency of the event and n is the size of the sample space.

Example 33.8Personality types are broadly defined according to four main preferences. Domarried couples choose similar or different personality types in their mates?The following data give an indication:

Similarities and Differences in a Random Sample of 545 Married Couples

Number of Similar Preferences Number of Married CouplesAll four 108Three 59Two 146One 141None 91

Suppose that a married couple is selected at random. Use the data to esti-mate the probability (to two decimal places) that they will have no person-ality preferences in common.

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Solution.The probability they will have no personality preferences in common isP (0) = 91

545= 0.17

Next, we define the probability of nonoccurrence of an event E (called itsfailure) to be the number P (Ec). Not surprisingly, the probabilities of anevent E and its complement Ec are related. The probability of the event Ec

is easily found from the identity

number of outcomes in A

total number of outcomes+

number of outcomes not in A

total number of outcomes= 1,

so that P (E) + P (Ec) = 1.

Example 33.9The probability that a college student without a flu shot will get the flu is0.45. What is the probability that a college student without the flu shot willnot get the flu?

Solution.Let E denote the event with outcomes those students without a flu shot.Then P (E) = 0.45. The probability that a student without the flu shot willnot get the flu is then P (Ec) = 1− P (E) = 1− 0.45 = 0.55.

Practice Problems

Problem 33.1An experiment consists of flipping a fair coin twice and recording each flip.Determine its sample space.

Problem 33.2Three coins are thrown. List the outcomes which belong to each of the fol-lowing events.

(a) exactly two tails (b) at least two tails (c) at most two tails.

Problem 33.3For each of the following events A, B, C, list and count the number of out-comes it contains and hence calculate the probability of A, B or C occurring.

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(a) A = ”throwing 3 or higher with one die”,(b) B = ”throwing exactly two heads with three coins”,(c) C = ”throwing a total score of 14 with two dice”.

Problem 33.4An experiment consists of throwing two four-faced dice.

(a) Write down the sample space of this experiment.(b) If E is the event total score is at least 4 list the outcomes belonging toEc.(c) If each die is fair find the probability that the total score is at least 6when the two dice are thrown. What is the probability that the total scoreis less than 6?(d) What is the probability that a double: (i.e. {(1, 1), (2, 2), (3, 3), (4, 4)})will not be thrown?(e) What is the probability that a double is not thrown nor is the scoregreater than 6?

Problem 33.5A lot consists of 10 good articles, 4 with minor defects and 2 with majordefects. One article is chosen at random. Find the probability that:

(a) it has no defects,(b) it has no major defects,(c) it is either good or has major defects.

Problem 33.6Consider the experiment of spinning the pointer on the game spinner pic-tured below. There are three possible outcomes, that is, when the pointerstops it must point to one of the three colors. (We rule out the possibility oflanding on the border between two colors.)

(a) What is the probability that the spinner is pointing to the red area?(b) What is the probability that the spinner is pointing to the blue area?(c) What is the probability that the spinner is pointing to the green area?

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Problem 33.7Consider the experiment of flipping a coin three times. If we denote a head byH and a tail by T, we can list the 8 possible ordered outcomes as (H,H,H),(H,H,T) each of which occurs with probability of 1/8. Finish listing theremaining members of the sample space. Calculate the probability of thefollowing events:

(a) All three flips are heads.(b) Exactly two flips are heads.(c) The first flip is tail.(d) At least one flip is head.

Problem 33.8Suppose an experimet consists of drawing one slip of paper from a jar con-taining 12 slips of paper, each with a different month of the year written onit. Find each of the following:

(a) The sample space S of the experiment.(b) The event A consisting of the outcomes having a month begining with J.(c) The event B consisting of outcomes having the name of a month that hasexactly four letters.(d) The event C consisting of outcomes having a month that begins with Mor N.

Problem 33.9Let S = {1, 2, 3, · · · , 25}. If a number is chosen at random, that is, with thesame chance of being drawn as all other numbers in the set, calculate each

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of the following probabilities:

(a) The even A that an even number is drawn.(b) The event B that a number less than 10 and greater than 20 is drawn.(c) The event C that a number less than 26 is drawn.(d) The event D that a prime number is drawn.(e) The event E that a number both even and prime is drawn.

Problem 33.10Consider the experiment of drawing a single card from a standard deck ofcards and determine which of the following are sample spaces with equallylikely outcomes:(a) {face card, not face card}(b) {club, diamond, heart, spade}(c) {black, red}(d) {king, queen, jack, ace, even card, odd card}

Problem 33.11An experiment consists of selecting the last digit of a telephone number.Assume that each of the 10 digits is equally likely to appear as a last digit.List each of the following:(a) The sample space(b) The event consisting of outcomes that the digit is less than 5(c) The event consisting of outcomes that the digit is odd(d) The event consisting of outcomes that the digit is not 2(e) Find the probability of each of the events in (b) - (d)

Problem 33.12Each letter of the alphabet is written on a seperate piece of paper and placedin a box and then one piece is drawn at random.(a) What is the probability that the selected piece of paper has a vowel writ-ten on it?(b) What is the probability that it has a consonant written on it?

Problem 33.13The following spinner is spun:

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Find the probabilities of obtaining each of the following:(a) P(factor of 35)(b) P(multiple of 3)(c) P(even number)(d) P(11)(e) P(composite number)(f) P(neither prime nor composite)

Problem 33.14An experiment consists of tossing four coins. List each of the following.

(a) The sample space(b) The event of a head on the first coin(c) The event of three heads

Problem 33.15Identify which of the following events are certain, impossible, or possible.(a) You throw a 2 on a die(b) A student in this class is less than 2 years old(c) Next week has only 5 days

Problem 33.16Two dice are thrown. If each face is equally likely to turn up, find thefollowing probabilities.(a) The sum is even(b) The sum is not 10(c) The sum is a prime(d) The sum is less than 9(e) The sum is not less than 9

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Problem 33.17What is the probability of getting yellow on each of the following spinners?

Problem 33.18A department store’s records show that 782 of 920 women who entered thestore on a Saturday afternoon made at least one purchase. Estimate theprobability that a woman who enters the store on a Saturday afternoon willmake at least one purchase.

Problem 33.19Suppose that a set of 10 rock samples includes 3 that contain gold nuggets.If you were to pick up a sample at random, what is the probability that itincludes a gold nugget?

Problem 33.20When do creative people get their best ideas? A magazine did a survey of 414inventors (who hold U.S. patents) and obtained the following information:

Time of Day When Best Ideas Occur

Time Number of Inventors6 A.M. - 12 noon 4612 noon - 6 P.M. 1886 P.M. - 12 midnight 6312 midnight - 6 A.M. 117

Assuming that the time interval includes the left limit and all the times upto but not including the right limit, estimate the probability (to two decimalplaces) that an inventor has a best idea during the time interval 6 A.M. - 12noon.

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Probability of Union and Intersection of Two EventsThe union of two events A and B is the event A ∪ B whose outcomes areeither in A or in B. The intersection of two events A and B is the eventA ∩ B whose outcomes are outcomes of both events A and B. Two eventsA and B are said to be mutually exclusive if they have no outcomes incommon. In this case A ∩B = ∅. Thus, P (A ∩B) = P (∅) = 0.

Example 33.10Consider the sample space of rolling a die. Let A be the event of rollingan even number, B the event of rolling an odd number, and C the event ofrolling a 2. Find

(a) A ∪B, A ∪ C, and B ∪ C.(b) A ∩B, A ∩ C, and B ∩ C.(c) Which events are mutually exclusive?


A ∪B = {1, 2, 3, 4, 5, 6}A ∪ C = {2, 4, 6}B ∪ C = {1, 2, 3, 5}

(b)A ∩B = ∅A ∩ C = {2}B ∩ C = ∅

(c) A and B are mutually exclusive as well as B and C.

Example 33.11Let A be the event of drawing a King from a well-shuffled standard deckof playing cards and B the event of drawing a ”ten” card. Are A and Bmutually exclusive?

Solution.Since A = {king of diamonds, king of hearts, king of clubs, king of spades}and B = {ten of diamonds, ten of hearts, ten of clubs, ten of spades} thenA and B are mutually exclusive since there are no cards common to bothevents.

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The next result provides a relationship between the probabilities of the eventsA, B, and A ∪B when A and B are mutually exclusive.

Theorem 33.1If A and B are mutually exclusive events of a sample space S than P (A∪B) =P (A) + P (B).

Proof.Since A and B are mutually exclusive then A and B have no elements incommon so that n(A ∪B) = n(A) + n(B). Thus,

P (A ∪B) =n(A ∪B)

n(S)=

n(A) + n(B)

n(S)=

n(A)

n(S)+

n(B)

n(S)= P (A) + P (B).

Now, for any events A and B, not necessarily exclusive, the probability ofA ∪B is given by the addition rule.

Theorem 33.2Let A and B be two events. Then

P (A ∪B) = P (A) + P (B)− P (A ∩B).

Proof.Let B −A denote the event whose outcomes are the outcomes in B that arenot in A. Then using the Venn diagram below we see that B = (A ∩ B) ∪(B − A) and A ∪B = A ∪ (B − A).

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Since (A ∩B) and (B − A) are disjoint then

P (B) = P (A ∩B) + P (B − A).

Thus,P (B − A) = P (B)− P (A ∩B).

Similarly, since A and B − A are disjoint then

P (A ∪B) = P (A) + P (B − A) = P (A) + P (B)− P (A ∩B).

Example 33.12Let P (A) = 0.9 and P (B) = 0.6. Find the minimum possible value forP (A ∩B).

Solution.Since P (A) + P (B) = 1.5 and 0 ≤ P (A ∪ B) ≤ 1 then by the previoustheorem

P (A ∩B) = P (A) + P (B)− P (A ∪B) ≥ 1.5− 1 = 0.5.

So the minimum value of P (A ∩B) is 0.5

Example 33.13Suppose there’s 40% chance of colder weather, 10% chance of rain and colderweather, 80% chance of rain or colder weather. Find the chance of rain.

Solution.By the addition rule we have

P (R) = P (R ∪ C)− P (C) + P (R ∩ C) = 0.8− 0.4 + 0.1 = 0.5.

Practice Problem

Problem 33.21Which of the following are mutually exclusive? Explain your answers.

(a) A driver getting a ticket for speeding and a ticket for going througha red light.(b) Being foreign-born and being President of the United States.

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Problem 33.22If A and B are the events that a consumer testing service will rate a givenstereo system very good or good, P (A) = 0.22, P (B) = 0.35. Find(a) P (Ac);(b) P (A ∪B);(c) P (A ∩B).

Problem 33.23If the probabilities are 0.20, 0.15, and 0.03 that a student will get a failinggrade in Statistics, in English, or in both, what is the probability that thestudent will get a failing grade in at least one of these subjects?

Problem 33.24If A is the event ”drawing an ace” from a deck of cards and B is the event”drawing a spade”. Are A and B mutually exclusive? Find P (A ∪B).

Problem 33.25A bag contains 18 coloured marbles: 4 are coloured red, 8 are colouredyellow and 6 are coloured green. A marble is selected at random. What isthe probability that the ball chosen is either red or green?

Problem 33.26Show that for any events A and B, P (A ∩B) ≥ P (A) + P (B)− 1.

Problem 33.27A golf bag contains 2 red tees, 4 blue tees, and 5 white tees.

(a) What is the probability of the event R that a tee drawn at randomis red?(b) What is the probability of the event ”not R” that is, that a tee drawn atrandom is not red?(c) What is the probability of the event that a tee drawn at random is eitherred or blue?

Problem 33.28A fair pair of dice is rolled. Let E be the event of rolling a sum that is aneven number and P the event of rolling a sum that is a prime number. Findthe probability of rolling a sum that is even or prime?

Problem 33.29If events A nd B are from the same sample space, and if P(A)=0.8 andP(B)=0.9, can events A and B be mutually exclusive?

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34 Probability and Counting Techniques

If you recall that the classical probability of an event E ⊆ S is given by

P (E) =n(E)

n(S)

where n(E) and n(S) denote the number of elements of E and S respectively.Thus, finding P (E) requires counting the elements of the sample space S.Sometimes the sample space is so large that shortcuts are needed to countall the possibilities.All the examples discussed thus far have been experiments consisting of oneaction such as tossing three coins or rolling two dice. We now want to con-sider experiments that consist of doing two or more actions in succession.For example, consider the experiment of drawing two balls in succession andwith replacement from a box containing one red ball (R), one white ball (W),and one green ball (G). The outcomes of this experiment, i.e. the elementsof the sample space can be found in two different ways by using

Organized Table or an Orderly ListAn organized table of our experiment looks like

R W GR RR RW RGW WR WW WGG GR GW GG

Thus, there are nine equally likely outcomes so that

S = {RR, RW, RG, WR,WW, WG, GR, GW, GG}

Tree DiagramsAn alternative way to generate the sample space is to use a tree diagram asshown in Figure 34.1.

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Figure 34.1

Example 34.1Show the sample space for tossing one penny and rolling one die. (H = heads,T = tails)

Solution.According to Figure 34.2, the sample space is

{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

Figure 34.2

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Fundamental Principle of CountingIf there are many stages to an experiment and several possibilities at eachstage, the tree diagram associated with the experiment would become toolarge to be manageable. For such problems the counting of the outcomes issimplified by means of algebraic formulas. The commonly used formula isthe multiplication rule of counting which states:” If a choice consists of k steps, of which the first can be made in n1 ways, foreach of these the second can be made in n2 ways,... and for each of these thekth can be made in nk ways, then the whole choice can be made in n1 ·n2 ·...nk

ways.”

Example 34.2How many license-plates with 3 letters followed by 3 digits exist?

Solution.A 6-step process: (1) Choose the first letter, (2) choose the second letter,(3) choose the third letter, (4) choose the first digit, (5) choose the seconddigit, and (6) choose the third digit. Every step can be done in a number ofways that does not depend on previous choices, and each license plate canbe specified in this manner. So there are 26 · 26 · 26 · 10 · 10 · 10 = 17, 576, 000ways.

Example 34.3How many numbers in the range 1000 - 9999 have no repeated digits?

Solution.A 4-step process: (1) Choose first digit, (2) choose second digit, (3) choosethird digit, (4) choose fourth digit. Every step can be done in a numberof ways that does not depend on previous choices, and each number can bespecified in this manner. So there are 9 · 9 · 8 · 7 = 4, 536 ways.

Example 34.4How many license-plates with 3 letters followed by 3 digits exist if exactlyone of the digits is 1?

Solution.In this case, we must pick a place for the 1 digit, and then the remainingdigit places must be populated from the digits {0, 2, · · · 9}. A 6-step process:(1) Choose the first letter, (2) choose the second letter, (3) choose the third

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letter, (4) choose which of three positions the 1 goes, (5) choose the firstof the other digits, and (6) choose the second of the other digits. Everystep can be done in a number of ways that does not depend on previouschoices, and each license plate can be specified in this manner. So there are26 · 26 · 26 · 3 · 9 · 9 = 4, 270, 968 ways.

Practice Problems

Problem 34.1If each of the 10 digits is chosen at random, how many ways can you choosethe following numbers?

(a) A two-digit code number, repeated digits permitted.(b) A three-digit identification card number, for which the first digit cannotbe a 0.(c) A four-digit bicycle lock number, where no digit can be used twice.(d) A five-digit zip code number, with the first digit not zero.

Problem 34.2(a) If eight horses are entered in a race and three finishing places are consid-ered, how many finishing orders can they finish?(b) If the top three horses are Lucky one, Lucky Two, and Lucky Three, inhow many possible orders can they finish?

Problem 34.3You are taking 3 shirts(red, blue, yellow) and 2 pairs of pants (tan, gray) ona trip. How many different choices of outfits do you have?

Problem 34.4The state of Maryland has automobile license plates consisting of 3 lettersfollowed by three digits. How many possible license plates are there?

Problem 34.5A club has 10 members. In how many ways can the club choose a presidentand vice-president if everyone is eligible?

Problem 34.6A lottery allows you to select a two-digit number. Each digit may be either1,2 or 3. Use a tree diagram to show the sample space and tell how manydifferent numbers can be selected.

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Problem 34.7In a medical study, patients are classified according to whether they haveblood type A, B, AB, or O, and also according to whether their blood pressureis low, normal, or high. Use a tree diagram to represent the various outcomesthat can occur.

Problem 34.8If a travel agency offers special weekend trips to 12 different cities, by air,rail, or bus, in how many different ways can such a trip be arranged?

Problem 34.9If twenty paintings are entered in art show, in how many different ways canthe judges award a first prize and a second prize?

Problem 34.10In how many ways can the 52 members of a labor union choose a president,a vice-president, a secretary, and a treasurer?

Problem 34.11Find the number of ways in which four of ten new movies can be ranked first,second, third, and fourth according to their attendance figures for the firstsix months.

Problem 34.12To fill a number of vacancies, the personnel manager of a company has tochoose three secretaries from among ten applicants and two bookkeepersfrom among five applicants. In how many different ways can the personnelmanager fill the five vacancies?

Finding Probabilities Using the Fundamental Principle of CountingThe Fundamental Principle of Counting can be used to compute probabilitiesas shown in the following example.

Example 34.5A quizz has 5 multiple-choice questions. Each question has 4 answer choices,of which 1 is correct answer and the other 3 are incorrect. Suppose that youguess all the answers.

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(a) How many ways are there to answer the 5 questions?(b) What is the probability of getting all 5 questions right?(c) What is the probability of getting exactly 4 questions right and 1 wrong?(d) What is the probability of doing well (getting at leat 4 right)?

Solution.(a) We can look at this question as a decision consisting of five steps. Thereare 4 ways to do each step so that by the Fundamental Principle of Countingthere are

(4)(4)(4)(4)(4) = 1024 ways

(b) There is only one way to answer each question correctly. Using theFundamental Principle of Counting there is (1)(1)(1)(1)(1) = 1 way to answerall 5 questions correctly out of 1024 possibilities. Hence,

P(all 5 right) = 11024

(c) The following table lists all possible responses that involve at least 4 rightanswers, R stands for right and W stands for a wrong answer

Five Responses Number of ways to fill out the testWRRRR (3)(1)(1)(1)(1) = 3RWRRR (1)(3)(1)(1)(1) = 3RRWRR (1)(1)(3)(1)(1) = 3RRRWR (1)(1)(1)(3)(1) = 3RRRRW (1)(1)(1)(1)(3) = 3

So there are 15 ways out of the 1024 possible ways that result in 4 rightanswers and 1 wrong answer so that

P(4 right,1 wrong) = 151024

≈ 1.5%

(d) ”At least 4” means you can get either 4 right and 1 wrong or all 5 right.Thus,

P(at least 4 right) = P(4 right, 1 wrong) + P(5 right) =15

1024+ 1

1024= 16

1024≈ 0.016

Probability TreesProbability trees can be used to compute the probabilities of combined out-comes in a sequence of experiments.

137

Example 34.6Construct the probability tree of the experiment of flipping a fair coin twice.

Solution.The probability tree is shown in Figure 34.3

Figure 34.3

The probabilities shown in Figure 34.3 are obtained by following the pathsleading to each of the four outcomes and multiplying the probabilities alongthe paths. This procedure is an instance of the following general property

Multiplication Rule for Probabilities for Tree DiagramsFor all multistage experiments, the probability of the outcome along anypath of a tree diagram is equal to the product of all the probabilities alongthe path.

Example 34.7Suppose that out of 500 computer chips there are 9 defective. Construct theprobability tree of the experiment of sampling two of them without replace-ment.

Solution.The probability tree is shown in Figure 34.4

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Figure 34.4

Practice Problems

Problem 34.13A box contains three red balls and two blue balls. Two balls are to be drawnwithout replacement. Use a tree diagram to represent the various outcomesthat can occur. What is the probability of each outcome?

Problem 34.14Repeat the previous exercise but this time replace the first ball before drawingthe second.

Problem 34.15If a new-car buyer has the choice of four body styles, three engines, and tencolors, in how many different ways can s/he order one of these cars?

Problem 34.16A jar contains three red gumballs and two green gumballs. An experimentconsists of drawing gumballs one at a time from the jar, without replacement,until a red one is obtained. Find the probability of the following events.

A: Only one draw is needed.B: Exactly two draws are needed.

C: Exactly three draws are needed.

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Problem 34.17Consider a jar with three black marbles and one red marble. For the exper-iment of drawing two marbles with replacement, what is the probability ofdrawing a black marble and then a red marble in that order?

Problem 34.18A jar contains three marbles, two black and one red. Two marbles are drawnwith replacement. What is the probability that both marbles are black?Assume that the marbles are equally likely to be drawn.

Problem 34.19A jar contains four marbles-one red, one green, one yellow, and one white.If two marbles are drawn without replacement from the jar, what is theprobability of getting a red marble and a white marble?

Problem 34.20A jar contains 3 white balls and 2 red balls. A ball is drawn at random fromthe box and not replaced. Then a second ball is drawn from the box. Draw atree diagram for this experiment and find the probability that the two ballsare of different colors.

Problem 34.21Suppose that a ball is drawn from the box in the previous problem, its colorrecorded, and then it is put back in the box. Draw a tree diagram for thisexperiment and find the probability that the two balls are of different colors.

Problem 34.22Suppose there are 19 balls in an urn. They are identical except in color. 16of the balls are black and 3 are purple. You are instructed to draw out oneball, note its color, and set it aside. Then you are to draw out another balland note its color. What is the probability of drawing out a black on thefirst draw and a purple on the second?

Binary ExperimentsBinary experiments are experiments with exactly two outcomes such as coin-tossing. Our first question is to find the total number of outcomes when acoin is tossed n times which is equivalent to saying that n coins are tossed.So what we have here is a decision consisting of n steps each step has twooutcomes (head/tail) so by the Fundamental Principle of Counting there are

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2 · 2 · 2 · · · 2︸︷︷︸n factors

= 2n outcomes

Next, let’s list the outcomes. For simplicity, assume n = 3. In this case, thetree diagram of Figure 34.5 lists all the outcomes of tossing one, two, andthree coins.

Figure 34.5

From Figure 34.5 we can create a tree diagram that counts the coin outcomeswith a given number of heads as shown in Figure 34.6.

Figure 34.6

The pattern of Figure 34.6 holds for any number of coins and thus leads tothe following diagram known as Pascal’s triangle.

141

Example 34.8Six fair coins are tossed.

(a) Find the probability of getting exactly 3 heads.(b) Find the probability of getting at least four heads.

Solution.(a) The 6-coins row of Pascal’s triangle may be interpreted as follows

1(6H) 6(5H) 15(4H) 20(3H) 15(2H) 6(1H) 1(0H)

Thus, there are 20 ways of getting exactly three heads, and the probabilityof 3 heads is 20

64= 5

16

(b) The first numbers −1, 6, and 15− represent the number of outcomes forwhich there are at least 4 heads. Thus, the probability of getting at leastfour heads is

1 + 6 + 15

64=

22

64=

11

32

Practice Problems

Problem 34.23The row of Pascal’s triangle that starts 1,4,... would be useful in findingprobabilities for an experiment of tossing four coins.(a) Interpret the meaning of each number.(b) Find the probability of exactly one head and three tails.(c) Find the probability of at least one tail turning up.

Problem 34.24Four coins are tossed.

(a) Draw a tree diagram to represent the arrangements of heads (H) and

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tails (T).(b) How many outcomes involve all heads? three heads, one tail? two heads,two tails? one head, three tails? no heads?(c) How do these results relate to Pascal’s triangle?

Problem 34.25A true-false problem has 6 questions.

(a) How many ways are there to answer the 6-question test?(b) What is the probability of getting at least 5 right by guessing the answersat random?

Problem 34.26(a) Write the 7th row of Pascal’s triangle.(b) What is the probability of getting at least four heads when tossing sevencoins?

Problem 34.27Assume the probability is 1

2that a child born is a boy. What is the probability

that if a family is going to have four children, they will all be boys?

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35 Permutations, Combinations and Proba-

bility

Thus far we have been able to list the elements of a sample space by drawinga tree diagram. For large sample spaces tree diagrams become very complexto construct. In this section we discuss counting techniques for finding thenumber of elements of a sample space or an event without having to list them.

PermutationsConsider the following problem: In how many ways can 8 horses finish in arace (assuming there are no ties)? We can look at this problem as a decisionconsisting of 8 steps. The first step is the possibility of a horse to finishfirst in the race, the second step the horse finishes second, ... , the 8th stepthe horse finishes 8th in the race. Thus, by the Fundamental Principle ofcounting there are

8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40, 320 ways

This problem exhibits an example of an ordered arrangement, that is, theorder the objects are arranged is important. Such ordered arrangement iscalled a permutation. Products such as 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 can be writtenin a shorthand notation called factoriel. That is, 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 8!(read ”8 factoriel”). In general, we define n factoriel by

n! =

{n(n− 1)(n− 2) · · · 3 · 2 · 1, if n ≥ 1

1, if n = 0.

where n is a whole number n.

Example 35.1Evaluate the following expressions:

(a) 6! (b) 10!7!

.

Solution.(a) 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720(b) 10!

7!= 10·9·8·7·6·5·4·3·2·1

7·6·5·4·3·2·1 = 10 · 9 · 8 = 720

Using factoriels we see that the number of permutations of n objects is n!.

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Example 35.2There are 6! permutations of the 6 letters of the word ”square.” In how manyof them is r the second letter?

Solution.Let r be the second letter. Then there are 5 ways to fill the first spot, 4ways to fill the third, 3 to fill the fourth, and so on. There are 5! suchpermutations.

Example 35.3Five different books are on a shelf. In how many different ways could youarrange them?

Solution.The five books can be arranged in 5 · 4 · 3 · 2 · 1 = 5! = 120 ways.

Counting PermutationsWe next consider the permutations of a set of objects taken from a largerset. Suppose we have n items. How many ordered arrangements of r itemscan we form from these n items? The number of permutations is denotedby P (n, r). The n refers to the number of different items and the r refers tothe number of them appearing in each arrangement. This is equivalent tofinding how many different ordered arrangements of people we can get on rchairs if we have n people to choose from. We proceed as follows.The first chair can be filled by any of the n people; the second by any of theremaining (n−1) people and so on. The rth chair can be filled by (n− r+1)people. Hence we easily see that

P (n, r) = n(n− 1)(n− 2)...(n− r + 1) =n!

(n− r)!.

Example 35.4How many ways can gold, silver, and bronze medals be awarded for a racerun by 8 people?

Solution.Using the permuation formula we find P (8, 3) = 8!

(8−3)!= 336 ways.

Example 35.5How many five-digit zip codes can be made where all digits are unique? Thepossible digits are the numbers 0 through 9.

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Solution.P (10, 5) = 10!

(10−5)!= 30, 240 zip codes.

Practice Problems

Problem 35.1Compute each of the following expressions.

(a) (2!)(3!)(4!)(b) (4× 3)!(c) 4 · 3!(d) 4!− 3!(e) 8!

5!

(f) 8!0!

Problem 35.2Compute each of the following.

(a) P (7, 2) (b) P (8, 8) (c) P (25, 2)

Problem 35.3Find m and n so that P (m, n) = 9!

6!

Problem 35.4How many four-letter code words can be formed using a standard 26-letteralphabet(a) if repetition is allowed?(b) if repetition is not allowed?

Problem 35.5Certain automobile license plates consist of a sequence of three letters fol-lowed by three digits.

(a) If no repetitions of letters are permitted, how many possible license platesare there?(b) If no letters and no digits are repeated, how many license plates arepossible?

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Problem 35.6A combination lock has 40 numbers on it.

(a) How many different three-number combinations can be made?(b) How many different combinations are there if the numbers must be alldifferent?

Problem 35.7(a) Miss Murphy wants to seat 12 of her students in a row for a class picture.How many different seating arrangements are there?(b) Seven of Miss Murphy’s students are girls and 5 are boys. In how manydifferent ways can she seat the 7 girls together on the left, and then the 5boys together on the right?

Problem 35.8Using the digits 1, 3, 5, 7, and 9, with no repetitions of the digits, how many

(a) one-digit numbers can be made?(b) two-digit numbers can be made?(c) three-digit numbers can be made?(d) four-digit numbers can be made?

Problem 35.9There are five members of the Math Club. In how many ways can thepositions of officers, a president and a treasurer, be chosen?

Problem 35.10(a) A baseball team has nine players. Find the number of ways the managercan arrange the batting order.(b) Find the number of ways of choosing three initials from the alphabet ifnone of the letters can be repeated.

CombinationsAs mentioned above, in a permutation the order of the set of objects or peo-ple is taken into account. However, there are many problems in which wewant to know the number of ways in which r objects can be selected fromn distinct objects in arbitrary order. For example, when selecting a two-person committee from a club of 10 members the order in the committee isirrelevant. That is choosing Mr A and Ms B in a committee is the same as

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choosing Ms B and Mr A. A combination is a group of items in which theorder does not make a difference.

Counting CombinationsLet C(n, r) denote the number of ways in which r objects can be selectedfrom a set of n distinct objects. Since the number of groups of r elementsout of n elements is C(n, r) and each group can be arranged in r! ways thenP (n, r) = r!C(n, r). It follows that

C(n, r) =P (n, r)

r!=

n!

r!(n− r)!.

Example 35.6How many ways can two slices of pizza be chosen from a plate containingone slice each of pepperoni, sausage, mushroom, and cheese pizza.

Solution.In choosing the slices of pizza, order is not important. This arrangement isa combination. Thus, we need to find C(4, 2) = 4!

2!(4−2)!= 6. So, there are six

ways to choose two slices of pizza from the plate.

Example 35.7How many ways are there to select a committee to develop a discrete mathe-matics course at a school if the committee is to consist of 3 faculty membersfrom the mathematics department and 4 from the computer science depart-ment, if there are 9 faculty members of the math department and 11 of theCS department?

Solution.There are C(9, 3) · C(11, 4) = 9!

3!(9−3)!· 11!

4!(11−4)!= 27, 720 ways.

Practice Problems

Problem 35.11Compute each of the following: (a) C(7,2) (b) C(8,8) (c) C(25,2)

Problem 35.12Find m and n so that C(m, n) = 13

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Problem 35.13The Library of Science Book Club offers three books from a list of 42. Ifyou circle three choices from a list of 42 numbers on a postcard, how manypossible choices are there?

Problem 35.14At the beginning of the second quarter of a mathematics class for elementaryschool teachers, each of the class’s 25 students shook hands with each of theother students exactly once. How many handshakes took place?

Problem 35.15There are five members of the math club. In how many ways can the two-person Social Committee be chosen?

Problem 35.16A consumer group plans to select 2 televisions from a shipment of 8 to checkthe picture quality. In how many ways can they choose 2 televisions?

Problem 35.17The Chess Club has six members. In how many ways(a) can all six members line up for a picture?(b) can they choose a president and a secretary?(c) can they choose three members to attend a regional tournament with noregard to order?

Problem 35.18Find the smallest value m and n such that C(m, n) = P (15, 2)

Problem 35.19A school has 30 teachers. In how many ways can the school choose 3 peopleto attend a national meeting?

Problem 35.20Which is usually greater the number of combinations of a set of objects orthe number of permutations?

Problem 35.21How many different 12-person juries can be chosen from a pool of 20 juries?

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Finding Probabilities Using Combinations and PermutationsCombinations can be used in finding probabilities as illustrated in the nextexample.

Example 35.8Given a class of 12 girls and 10 boys.

(a) In how many ways can a committee of five consisting of 3 girls and 2boys be chosen?(b) What is the probability that a committee of five, chosen at random fromthe class, consists of three girls and two boys?(c) How many of the possible committees of five have no boys?(i.e. consistsonly of girls)(d) What is the probability that a committee of five, chosen at random fromthe class, consists only of girls?

Solution.(a) First note that the order of the children in the committee does not matter.From 12 girls we can choose C(12, 3) different groups of three girls. From the10 boys we can choose C(10, 2) different groups. Thus, by the FundamentalPrinciple of Counting the total number of committee is

C(12, 3) · C(10, 2) =12!

3!9!· 10!

2!8!=

12 · 11 · 10

3 · 2 · 1· 10 · 9

2 · 1= 9900

(b) The total number of committees of 5 is C(22, 5) = 26, 334. Using part(a), we find the probability that a committee of five will consist of 3 girls and2 boys to be

C(12, 3) · C(10, 2)

C(22, 5)=

9900

26, 334≈ 0.3759.

(c) The number of ways to choose 5 girls from the 12 girls in the class is

C(10, 0) · C(12, 5) = C(12, 5) =12 · 11 · 10 · 9 · 8

5 · 4 · 3 · 2 · 1= 792

(d) The probability that a committee of five consists only of girls is

C(12, 5)

C(22, 5)=

792

26, 334≈ 0.03

150

Practice Problems

Problem 35.22John and Beth are hoping to be selected from their class of 30 as presidentand vice-president of the Social Committee. If the three-person committee(president, vice-president, and secretary) is selected at random, what is theprobability that John and Beth would be president and vice president?

Problem 35.23There are 10 boys and 13 girls in Mr. Benson’s fourth-grade class and 12boys and 11 girls in Mr. Johnson fourth-grade class. A picnic committeeof six people is selected at random from the total group of students in bothclasses.

(a) What is the probability that all the committee members are girls?(b) What is the probability that the committee has three girls and threeboys?

Problem 35.24A school dance committee of 4 people is selected at random from a group of6 ninth graders, 11 eighth graders, and 10 seventh graders.

(a) What is the probability that the committee has all seventh graders?(b) What is the probability that the committee has no seventh graders?

Problem 35.25In an effort to promote school spirit, Georgetown High School created IDnumbers with just the letters G, H, and S. If each letter is used exactly threetimes,

(a) how many nine-letter ID numbers can be generated?(b) what is the probability that a random ID number starts with GHS?

Problem 35.26The license plates in the state of Utah consist of three letters followed bythree single-digit numbers.

(a) If Edward’s initials are EAM, what is the probability that his licenseplate will have his initials on it (in any order)?

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(b) What is the probability that his license plate will have his initials in thecorrect order?

152

36 Odds, Expected Value, and Conditional

Probability

What’s the difference between probabilities and odds? To answer this ques-tion, let’s consider a game that involves rolling a die. If one gets the face1 then he wins the game, otherwise he loses. The probability of winningis 1

6whereas the probability of losing is 5

6. The odds of winning is 1:5(read

1 to 5). This expression means that the probability of losing is five timesthe probability of winning. Thus, probabilities describe the frequency of afavorable result in relation to all possible outcomes whereas the ”odds infavor” compare the favorable outcomes to the unfavorable outcomes. Moreformally,

odds in favor = favorable outcomesunfavorable outcomes

If E is the event of all favorable outcomes then its complementary, E, is theevent of unfavorable outcomes. Hence,

odds in favor = n(E)

n(E)

Also, we define the odds against an event as

odds against = unfavorable outcomesfavorable outcomes

= n(E)n(E)

Any probability can be converted to odds, and any odds can be converted toa probability.

Converting Odds to ProbabilitySuppose that the odds for an event E is a:b. Thus, n(E) = ak and n(E) = bkwhere k is a positive integer. Since E and E are complementary then

153

n(S) = n(E) + n(E). Therefore,

P (E) = n(E)n(S)

= n(E)

n(E)+n(E)

= akak+bk

= aa+b

P (E) = n(E)n(S)

= n(E)

n(E)+n(E)

= bkak+bk

= ba+b

Example 36.1If the odds in favor of an event E is 5 to 4, compute P (E) and P (E).

Solution.We have

P (E) =5

5 + 4=

5

9

and

P (E) =4

5 + 4=

4

9

Converting Probability to OddsGiven P (E), we want to find the odds in favor of E and the odds against E.The odds in favor of E are

n(E)

n(E)= n(E)

n(S)· n(S)

n(E)

= P (E)

P (E)

= P (E)1−P (E)

and the odds against E are

n(E)

n(E)=

1− P (E)

P (E)

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Example 36.2For each of the following, find the odds in favor of the event’s occurring:

(a) Rolling a number less than 5 on a die.(b) Tossing heads on a fair coin.(c) Drawing an ace from an ordinary 52-card deck.

Solution.(a) The probability of rolling a number less than 5 is 4

6and that of rolling 5

or 6 is 26. Thus, the odds in favor of rolling a number less than 5 is 4

6÷ 2

6= 2

1

or 2:1(b) Since P (H) = 1

2and P (T ) = 1

2then the odds in favor of getting heads is(

12

)÷

(12

)or 1:1

(c) We have P(ace) = 452

and P(not an ace) =4852

so that the odds in favor ofdrawing an ace is

(452

)÷

(4852

)= 1

12or 1:12

Remark 36.1A probability such as P (E) = 5

6is just a ratio. The exact number of favorable

outcomes and the exact total of all outcomes are not necessarily known.

Practice Problems

Problem 36.1If the probability of a boy’s being born is 1

2, and a family plans to have four

children, what are the odds against having all boys?

Problem 36.2If the odds against Deborah’s winning first prize in a chess tournament are3 to 5, what is the probability that she will win first prize?

Problem 36.3What are the odds in favor of getting at least two heads if a fair coin is tossedthree times?

Problem 36.4If the probability of rain for the day is 60%, what are the odds against itsraining?

155

Problem 36.5On a tote board at a race track, the odds for Gameylegs are listed as 26:1.Tote boards list the odds that the horse will lose the race. If this is the case,what is the probability of Gameylegs’s winning the race?

Problem 36.6If a die is tossed, what are the odds in favor of the following events?(a) Getting a 4(b) Getting a prime(c) Getting a number greater than 0(d) Getting a number greater than 6.

Problem 36.7Find the odds against E if P (E) = 3

4.

Problem 36.8Find P(E) in each case.

(a) The odds in favor of E are 3:4(b) The odds against E are 7:3

Expected ValueA cube has three red faces, two green faces, and one blue face. A gameconsists of rolling the cube twice. You pay $ 2 to play. If both faces are thesame color, you are paid $ 5(that is you win $3). If not, you lose the $2 itcosts to play. Will you win money in the long run? Let W denote the eventthat you win. Then W = {RR, GG, BB} and

P (W ) = P (RR) + P (GG) + P (BB) =1

2· 1

2+

1

3· 1

3+

1

6· 1

6=

7

18≈ 39%.

Thus, P (L) = 1118

= 61%. Hence, if you play the game 18 times you expectto win 7 times and lose 11 times on average. So your winnings in dollars willbe 3 × 7 − 2 × 11 = −1. That is, you can expect to lose $1 if you play thegame 18 times. On the average, you will lose 1

18per game (about 6 cents).

This can be found also using the equation

3× 7

18− 2× 11

18= − 1

18

156

We call this number the expected value. More formally, let the outcomesof an experiment be a sequence of real numbers n1, n2, · · · , nk, and supposethat the outcomes occur with respective probabilities p1, p2, · · · , pk. Then theexpected value of the experiment is

E = n1p1 + n2p2 + · · ·+ nkpk.

Example 36.3Suppose that an insurance company has broken down yearly automobileclaims for drivers from age 16 through 21 as shown in the following table.

Amount of claim Probability0 0.802,000 0.104,000 0.056,000 0.038,000 0.0110,000 0.01

How much should the company charge as its average premium in order tobreak even on costs for claims?

Solution.Finding the expected value

E = 0(0.80)+2, 000(0.10)+4, 000(0.05)+6, 000(0.03)+8, 000(0.01)+10, 000(0.01) = 760

Since average claim value is $760, the average automobile insurance premiumshould be set at $760 per year for the insurance company to break even

Example 36.4An American roulette wheel has 38 compartments around its rim. Two ofthese are colored green and are numbered 0 and 00. The remaining compart-ments are numbered 1 through 36 and are alternately colored black and red.When the wheel is spun in one direction, a small ivory ball is rolled in theopposite direction around the rim. When the wheel and the ball slow down,the ball eventually falls in any one compartments with equal likelyhood ifthe wheel is fair. One way to play is to bet on whether the ball will fall in ared slot or a black slot. If you bet on red for example, you win the amountof the bet if the ball lands in a red slot; otherwise you lose. What is theexpected win if you consistently bet $5 on red?

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Solution.The probability of winning is 18

38and that of losing is 20

38. Your expected win

is18

38× 5− 20

38× 5 ≈ −0.26

On average you should expect to lose 26 cents per play

Practice Problems

Problem 36.9Compute the expected value of the score when rolling two dice.

Problem 36.10A game consists of rolling two dice. You win the amounts shown for rollingthe score shown.

Score 2 3 4 5 6 7 8 9 10 11 12$ won 4 6 8 10 20 40 20 10 8 6 4

Compute the expected value of the game.

Problem 36.11Consider the spinner in Figure 36.1, with the payoff in each sector of thecircle. Should the owner of this spinner expect to make money over anextended period of time if the charge is $2.00 per spin?

Figure 36.1

Problem 36.12You play a game in which two dice are rolled. If a sum of 7 appears, youwin $10; otherwise, you lose $2.00. If you intend to play this game for along time, should you expect to make money, lose money, or come out abouteven? Explain.

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Problem 36.13Suppose it costs $8 to roll a pair of dice. You get paid the sum of the numbersin dollars that appear on the dice. What is the expected value of this game?

Problem 36.14An insurance company will insure your dorm room against theft for a semester.Suppose the value of your possessions is $800. The probability of your beingrobbed of $400 worth of goods during a semester is 1

100, and the probability

of your being robbed of $800 worth of goods is 1400

. Assume that these arethe only possible kinds of robberies. How much should the insurance com-pany charge people like you to cover the money they pay out and to makean additional $20 profit per person on the average?

Problem 36.15Consider a lottery game in which 7 out of 10 people lose, 1 out of 10 wins$50, and 2 out of 10 wins $35. If you played 10 times, about how much wouldyou expect to win?

Problem 36.16Suppose a lottery game allows you to select a 2-digit number. Each digitmay be either 1, 2, 3, 4, or 5. If you pick the winning number, you win $10.Otherwise, you win nothing. What is the expected payoff?

Conditional Probability and Independent EventsWhen the sample space of an experiment is affected by additional informa-tion, the new sample space is reduced in size. For example, suppose we tossa fair coin three times and consider the following events:

A : getting a tail on the first tossB : getting a tail on all three tosses

Since

S = {HHH, HHT,HTH,HTT, THH, THT, TTH, TTT}

then P (A) = 48

= 12

and P (B) = 18. What if we were told that event A has

occurred (that is, a tail occurred on the first toss), and we are now asked tofind P (B). The sample space is now reduced to {THH, THT, TTH, TTT}.The probability that all three are tails given that the first toss is a tail is 1

4.

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The notation we use for this situation is P (B|A), read ”the probability of Bgiven A,” and we write P (B|A) = 1

4. Notice that

P (B|A) =P (A ∩B)

P (A).

This is true in general, and we have the following:

Given two events A and B belonging to the same sample S. The condi-tional probability P (B|A) denotes the probability that event B will occurgiven that event A has occurred. It is given by the formula

P (B|A) =P (A ∩B)

P (A).

Example 36.5Consider the experiment of tossing a fair die. Denote by A and B the follow-ing events:

A = {Observing an even number of dots on the upper face of the die} ,B = {Observing a number of dots less than or equal to 3 on the upper face of the die}.

Find the probability of the event A, given the event B.

Solution.Since A = {2, 4, 6} and B = {1, 2, 3} then A ∩ B = {2} and therefore

P (A|B) =1636

= 13.

If P (B|A) = P (B), i.e., the occurrence of the event A does not affect theprobability of the event B, then we say that the two events A and B areindependent. In this case the above formula gives

P (A ∩B) = P (A) · P (B).

This formula is known as the ”multiplication rule of probabilities”. If twoevents are not independent, we say that they are dependent. In this case,P (B|A) 6= P (B) or equivalently P (A ∩B) 6= P (A) · P (B).

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Example 36.6Consider the experiment of tossing a fair die. Denote by A and B the follow-ing events:

A = {Observing an even number of dots on the upper face of the die} ,B = {Observing a number of dots less than or equal to 4 on the upper face of the die}.

Are A and B independent?

Solution.Since A = {2, 4, 6} and B = {1, 2, 3, 4} then A ∩ B = {2, 4} and therefore

P (A|B) =2646

= 12

= P (A). Thus, A and B are independent.

Practice Problems

Problem 36.17Suppose that A is the event of rolling a sum of 7 with two fair dice. Makeup an event B so that(a) A and B are independent.(b) A and B are dependent.

Problem 36.18When tossing three fair coins, what is the probability of getting two tailsgiven that the first coin came up heads?

Problem 36.19Suppose a 20-sided die has the following numerals on its face:1, 1, 2, 2, 2, 3,3, 4, 5, 6, 7, ,8 9, 10, 11, 12, 13, 14, 15, 16. The die is rolled once and thenumber on the top face is recorded. Let A be the event the number is prime,and B be the event the number is odd. Find P (A|B) and P (B|A).

Problem 36.20What is the probability of rolling a 6 on a fair die if you know that the rollis an even number?

Problem 36.21A red die and a green die are rolled. What is the probability of obtaining aneven number on the red die and a multiple of 3 on the green die?

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Problem 36.22Two coins are tossed. What is the probability of obtaining a head on thefirst coin and a tail on the second coin?

Problem 36.23Consider two boxes: Box 1 contains 2 white and 2 black balls, and box2 contains 2 white balls and three black balls. What is the probability ofdrawing a black ball from each box?

Problem 36.24A container holds three red balls and five blue balls. One ball is drawn anddiscarded. Then a second ball is drawn.(a) What is the probability that the second ball drawn is red if you drew ared ball the first time?(b) What is the probability of drawing a blue ball second if the first ball wasred?(c) What is the probability of drawing a blue ball second if the first ball wasblue?

Problem 36.25Consider the following events.

A: rain tomorrowB: You carry an umbrellaC: coin flipped tomorrow lands on heads

Which of two events are dependent and which are independent?

Problem 36.26You roll a regular red die and a regular green die. Consider the followingevents.

A: a 4 on the red dieB: a 3 on the green dieC: a sum of 9 on the two dice

Tell whether each pair of events is independent or dependent.(a) A and B (b) B and C

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37 Basic Geometric Shapes and Figures

In this section we discuss basic geometric shapes and figures such as points,lines, line segments, planes, angles, triangles, and quadrilaterals.

The three pillars of geometry are points, lines, and planes: A point is anundefined term used to describe for example a location on a map. A pointhas no length, width, or thickness and we often use a dot to represent it.Points are usually labeled with uppercase letters.Line is another basic term of geometry. Like a point, a line is an undefinedterm used to describe a tightly stretched thread or a laser beam. We cansay that a line is a straight arrangement of points. A line has no thicknessbut its length goes on forever in two directions as shown in Figure 37.1. Thearrows represent the fact that the line extends in both directions forever. Aline is often named by a lowercase letter such as the line k in Figure 37.1.

Figure 37.1

The subset of the line k consisting of all points between A and B togetherwith A and B forms a line segment denoted by AB. We call A the leftendpoint and B the right endpoint. The distance between the endpointsis known as the length of the line segment and will be denoted by AB. Twoline segments with the same length are said to be congruent.Any three or more points that belong to the same line are called collinear.See Figure 37.2

Figure 37.2

Three noncollinear points (also knwon as coplanar points) determine aplane, which is yet another undefined term used to describe a flat spacesuch as a tabletop.

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Subsets of a plane are called plane shapes or planes figures. We havealready discussed a geometric figure, namely, a line. Another importantexample of a geometric figure is the concept of an angle.By an angle we mean the opening between two line segments that have acommon endpoint, known as the vertex, as shown in Figure 37.3(a). Theline segments are called the sides of the angle.If one of the line segments of an angle is horizontal and the other is verticalthen we call the angle a right angle. See Figure 37.3(b). Note that thesides of an angle partition the plane into two regions, the interior and theexterior of the angle as shown in Figure 37.3(c). Two angles with the sameopening are said to be congruent.

Figure 37.3

The Early Stages of Learning GeometryThe first stage of a child’s learning geometry consists on recognizing geo-metric shapes by their appearances without paying attention to their compo-nent parts (such as the sides and the angles). For example, a rectangle maybe recognized because it ”looks like a door,” not because it has four straightsides and four right angles. The second stage, known as description, stu-dents are able to describe the component parts and properties of a shape,such as how many sides it has and whether it has some congruent sides orangles. At the third stage, students become aware of relationships betweendifferent shapes such as a rhombus is a quadrilateral with four congruentsides and a parallelogram is a quadrilateral with parallel opposite sides, etc.

TrianglesA triangle is a closed figure composed exactly of three line segments calledthe sides. The points of intersection of any two line segments is called avertex. Thus, a triangle has three vertices. Aslo, a triangle has three inte-rior angles. See Figure 37.4(a).Triangles may be classified according to their angles and sides. If exactly one

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of the angle is a right angle then the triangle is called a right triangle. SeeFigure 37.4(b). A triangle with three congruent sides is called an equilat-eral triangle. See Figure 37.4(c). A triangle with two or more congruentsides is called an isosceles triangle. A triangle with no congruent sides iscalled a scalene triangle.

Figure 37.4

QuadrilateralsBy a quadrilateral we mean a closed figure with exactly four line segments(or sides). Quadrilaterals are classified as follows:• A trapezoid is a quadrilateral that has exactly one pair of parallel sides.Model: the middle part of a bike frame.• An isosceles trapezoid is a quadrilateral with exactly two parallel sidesand the remaining two sides are congruent. Model: A water glass silhouette.• A parallelogram is a quadrilateral in which each pair of opposite sides isparallel.• A rhombus is a parallelogram that has four congruent sides. Model:diamond.• A kite is a quadrilateral with two nonoverlapping pairs of adjacent sidesthat are the same length. Model: a kite.• A rectangle is a parallelogram that has four right angles. Model: a door.• A square is a rectangle that has four congruent sides. Model: Floor tile.

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Figure 37.5

Practice Problems

Problem 37.1Find three objects in your classroom with surfaces that suggests commongeometric figures.

Problem 37.2A fifth grader says a square is not a rectangle because a square has fourcongruent sides and rectangles don’t have that. A second fifth grader saysa square is a type of rectangle because it is a parallelogram and it has fourright angles.

(a) Which child is right?(b) How can you use the definitions to help the other child understand?

Problem 37.3Suppose P={parallelograms}, Rh={rhombus}, S={squares}, Re={rectangles},T={trapezoids}, and Q={quadrilaterals}. Find(a) Rh ∩Re (b) T ∩ P

Problem 37.4Organize the sets P, Rh, S, Re, T, and Q using Venn diagram.

Problem 37.5(a) True or false? No scalene triangle is isosceles.(b) What shape is the diamond in a deck of cards?

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Problem 37.6How many squares are in the following design?

Problem 37.7Tell whether each of the following shapes must, can, or cannot have at leastone right angle.

(a) Rhombus(b) Square(c) Trapezoid(d) Rectangle(e) Parallelogram

Problem 37.8In which of the following shapes are both pairs of opposite sides parallel?


Problem 37.9A square is also which of the following?(a) Quadrilateral(b) Parallelogram(c) Rhombus(d) Rectangle

Problem 37.10Fill in the blank with ”All”, ”Some”, or ”No”

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(a) rectangles are squares.(b) parallelograms are trapezoids.(c) rhombuses are quadrilaterals.

Problem 37.11How many triangles are in the following design?

Problem 37.12How many squares are found in the following figure?

Problem 37.13Given are a variety of triangles. Sides with the same length are indicated.Right angles are indicated.

(a) Name the triangles that are scalene.(b) Name the triangles that are isosceles.(c) Name the triangles that are equilateral.(d) Name the triangles that contain a right angle.

Problem 37.14(a) How many triangles are in the figure?

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(b) How many parallelograms are in the figure?(c) How many trapezoids are in the figure?

Problem 37.15If possible, sketch two parallelograms that intersect at exactly(a) one point(b) two points(c) three points(d) four points.

Problem 37.16If possible, draw a triangle and a quadrilateral that intersect at exactly(a) one point(b) two points(c) three points.

Problem 37.17Suppose P={parallelograms}, S={squares}, T={trapezoids}, and Q={quadrilaterals}.Find

(a) P ∩ S(b) P ∪Q

Problem 37.18A fifth grader does not think that a rectangle is a type of parallelogram. Tellwhy it is.

Problem 37.19Tell whether each definition has sufficient information. If it is not sufficient,tell what information is missing.

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(a) A rhombus is a quadrilateral with both pairs of opposite sides paral-lel.(b) A square is a quadrilateral with four congruent sides.(c) A rhombus is a quadrilateral that has four congruent sides.

Problem 37.20Name properties that a square, parallelogram, and rhombus have in common.

Problem 37.21How many different line segments are contained in the following portion of aline?

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38 Properties of Geometric Shapes

In this section we study the properties of geometric shapes that belong tothe same plane such as lines and angles.

LinesTwo lines in a plane that do not have a point in common are called parallel.See Figure 38.1 (a). If the two lines are l and m then we write l||m. Whentwo or more lines have exactly one point in common then we say that theyare concurrent. See Figure 38.1(b).The midpoint of a line segment is a point that is equidistant from A andB, that is a point that has the same distance from A and B. See Figure38.1(c). A ray is a subset of a line that contains the endpoint and all pointson the line on one side of the point. See Figure 38.1(d).

Figure 38.1

Next, we list some of the important properties of lines.

Properties(1) Through any distinct two points there is exactly one line going throughthem.(2) If two points are in a plane then the line containing them lies in thatplane.(3) There is exactly one plane containing any three noncollinear distinctpoints.(4) A line and a point not on the line determine a plane.(5) Two parallel lines determine a plane.(6) Two concurrent lines determine a plane.

Angles: Measurements and TypesWe have seen that an angle is the opening between two line segments with

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one common endpoint. An angle can be also defined as the opening be-tween two rays with the same endpoint called the vertex. We call the raysthe sides of the angle. If A is the vertex, B is a point on one side and C isa point on the other side then we write ∠BAC. See Figure 38.2. Note thatthe middle letter in the notation of an angle is always the vertex. Sometimesonly the vertex A is given. In this case we use the notation ∠A. Also, anglescan be labeled using numbers. In this case we use the notation ∠1 to denotethe angle labeled 1.

An angle is measured according to the amount of ”opening” between itssides. A unit of angle measure is the degree. A complete rotation about apoint has a measure of 360 degrees, written 360◦. Thus, one degree is 1

360of

a complete rotation. We will write m(∠BAC) for the measure of the angleat vertex A.Angles are measured using a protractor. To measure an angle ∠BAC, placethe center of the protractor at the vertex A while lining up the side AB withthe base of the protractor. The other side of the angle will pass througha number on the protractor representing m(∠BAC) in degrees. See Figure38.2.

Figure 38.2

Next, we categorize angles according to their measures. If m(∠BAC) =90◦ then we will say that the angle is a right angle. See Figure 38.3(a).Two lines that intersect at a right angle are called perpendicular. If

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m(∠BAC) = 180◦ we say that the angle is straight. See Figure 38.3(b). If0◦ < m(∠BAC) < 90◦ then the angle is called acute. See Figure 38.3(c). If90◦ < m(∠BAC) < 180◦ then the angle is called obtuse. See Figure 38.3(d).If m(∠BAC) > 180◦ then the angle is called reflex. See Figure 38.3(e).

Figure 38.3

Now, recall that a triangle has three interior angles. If one of the angles is aright angle then we call the triangle a right triangle. If one of the angle isobtuse then we call the triangle an obtuse triangle. If all three angles areacute angles then the triangle is called an acute triangle.

Pairs of Angles and the Vertical Angles TheoremTwo angles that have a common side and nonoverlapping interiors are calledadjacent angles. See Figure 38.4(a). If the sum of two adjacent angles is90◦ then the two angles are called complementary. See Figure 38.4(b).Two adjacent angles whose sum is 180◦ are called supplementary angles.See Figure 38.4(c). A pair of nonadjacent angles formed by two intersectinglines are called vertical angles. For example, m(∠1) and m(∠2) are verticalangles as well as m(∠3) and m(∠4). See Figure 38.4(d)

Figure 38.4

Vertical angles are congruent as shown in the following theorem.

Theorem 38.1 (Vertical Angles Theorem)Vertical angles have the same measure.

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Proof.The angles ∠1 and ∠3 are supplementary so that m(∠1) + m(∠3) = 180◦.Similarly, the angles ∠2 and ∠3 are supplementary so that m(∠2)+m(∠3) =180◦. Thus, m(∠1)+m(∠3) = m(∠2)+m(∠3). Subtracting m(∠3) from bothsides of the equation gives m(∠1) = m(∠2). A similar argument shows thatm(∠3) = m(∠4)

Angles Associated with Parallel LinesIf two lines k and l are parallel then a line m as shown in Figure 38.5 iscalled a transversal line. In this case, there are 8 angles determined bythese three lines numbered 1 through 8.

Figure 38.5

Angles ∠1 and ∠6 as well as angles ∠4 and ∠5 are called alternate interiorangles. The following theorem shows that any pair of alternate interiorangles are congruent.

Theorem 38.2Any pair of alternate interior angles have the same measure.

Proof.We will show that m(∠4) = m(∠5). The other equality is similar. Accordingto Figure 38.6 we have the following equalities: m(∠c) + m(∠5) = 90◦ andm(∠4)+m(∠c) = 90◦. Thus, m(∠c)+m(∠5) = m(∠4)+m(∠c). Subtractingm(∠c) from both sides we obtain m(∠4) = m(∠5)

Figure 38.6

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Now, going back to Figure 38.5, the angles ∠1 and ∠8; ∠2 and ∠5; ∠3 and ∠6;∠4 and ∠7 are called corresponding angles. Every pair of correspondingangles are congruent as shown in the next theorem.

Theorem 38.3Corresponding angles have the same measure.

Proof.We will just show that m(∠2) = m(∠5). By the vertical angles theorem wehave m(∠2) = m(∠4). By the alternate interior angles theorem we havem(∠4) = m(∠5). From these two equalities we conclude that m(∠2) =m(∠5)

Sum of the Measures of the Interior Angles of a TriangleAs an application to the results of the previous paragraph, we show that thesum of the measures of the interior angles of a triangle is 180◦. To see this,let’s consider a triangle with vertices A, B, and C and interior angles ∠1, ∠2,and ∠3 as shown in Figure 38.7(a). At A draw a line parallel to the linesegment BC. Then using the corresponding angles theorem we can drawFigure 38.7(b) which shows that

m(∠1) + m(∠2) + m(∠3) = 180◦

as required.

Figure 38.7

Practice Problems

Problem 38.1Using Figure 38.5 show that m(∠1) + m(∠5) = 180◦.

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Problem 38.2(a) How many angles are shown in the following figure?

(b) How many are obtuse?(c) How many are acute?

Problem 38.3Find the missing angle in the following triangle.

Problem 38.4In the figure below, m(∠BFC) = 55◦, m(∠AFD) = 150◦, m(∠BFE) =120◦. Determine the measures of m(∠AFB) and m(∠CFD).

Problem 38.5In the following figure m(∠1) = m(∠2)

2− 9◦. Determine m(∠1) and m(∠2).

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Problem 38.6Angles 3 and 8; 2 and 7 in Figure 38.5 are called alternate exterior angles.Show that m(∠3) = m(∠8).

Problem 38.7Following are the measures of ∠A, ∠B, ∠C. Can a triangle ∆ABC be madethat has the given angles? Explain.

(a) m(∠A) = 36◦, m(∠B) = 78◦, m(∠C) = 66◦.(b) m(∠A) = 124◦, m(∠B) = 56◦, m(∠C) = 20◦.(c) m(∠A) = 90◦, m(∠B) = 74◦, m(∠C) = 18◦.

Problem 38.8In the following figure AO is perpendicular to CO. If m(∠AOD) = 165◦ andm(∠BOD) = 82◦, determine the measures of ∠AOB and ∠BOC.

Problem 38.9In the figure below, find the measures of ∠1, ∠2, ∠3, and ∠4.

Problem 38.10(a) How is a line segment different from a line?

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(b) What is the vertex of the angle ∠PAT?(c) How is AB different from AB?

Problem 38.11A fourth grader thinks that m(∠A) is greater than m(∠B).

(a) Why might the child think this?(b) How could you put the angles together to show that m(∠A) < m(∠B)?

Problem 38.12In the figure below

(a) name two supplementary angles(b) name two complementary angles.

Problem 38.13An angle measures 20◦. What is the measure of(a) its supplement? (b) its complement?

Problem 38.14True or false? If false give an example.

(a) All right angles are congruent.(b) Two complementary angles are congruent.(c) Two supplementary angles are congruent.

Problem 38.15Find the measures of the angles in the following figure.

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Problem 38.16How many degrees does the minute hand of a clock turn through

(a) in sixty minutes?(b) in five minutes?(c) in one minute?

Problem 38.17How many degrees does the hour hand of a clock turn through

(a) in sixty minutes?(b) in five minutes?

Problem 38.18Find the angle formed by the minute and hour hands of a clock at thesetimes.(a) 3:00 (b) 6:00 (c) 4:30 (d) 10:20

Problem 38.19Determine the measures of the interior angles.

Problem 38.20(a) Can a triangle have two obtuse angles? Why?(b) Can a triangle have two right angles? Why?(c) Can a triangle have two acute angles?

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Problem 38.21A hiker started heading due north, then turned to the right 38◦, then turnedto the left 57◦, and next turned right 9◦. To resume heading north, what turnmust be made?

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39 Symmetry of Plane Figures

In this section, we are interested in the symmetric properties of plane figures.By a symmetry of a plane figure we mean a motion of the plane that movesthe figure so that it falls back on itself.The two types of symmetry that we discuss are

Reflection Symmetry:A plane figure is symmetric about a line if it is its own image when flippedacross the line. We call the reflection line the line of symmetry. In otherwords, a figure has a line symmetry if it can be folded along the line so thatone half of the figure matches the other half.Reflection symmetry is also known as mirror symmetry, since the line ofsymmetry acts like a double-sided mirror. Points on each side are reflectedto the opposite side.Many plane figures have several line symmetries. Figure 39.1 shows some ofthe plane figures with their line symmetries shown dashed.

Figure 39.1

Remark 39.1Using reflection symmetry we can establish properties for some plane figures.For example, since an isosceles triangle has one line of symmetry then the

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base angles, i.e. angles opposed the congruent sides, are congruent. Asimilar property holds for isosceles trapezoids.

Rotational Symmetry:A plane figure has rotational symmetry if and only if it can be rotatedmore than 0◦ and less than or equal to 360◦ about a fixed point called thecenter of rotation so that its image coincides with its original position.Figure 39.2 shows the four different rotations of a square.

Figure 39.2

The following table list the number of turns, including a full turn, of someplane figures.

Figure RotationRectangle 2 rotations(180◦ and 360◦)Square 4 rotations(90◦, 180◦, 270◦, 360◦)Rhombus 2 rotations(180◦ and 360◦)Parallelogram 2 rotations(180◦ and 360◦)Trapezoid 1 rotation(360◦)Equilateral Triangle 3 rotations(120◦, 240◦, and 360◦)

Example 39.1Show that the opposite sides of a parallelogram are congruent.

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Solution.Rotate the parallelogram 180◦ about its center as shown in Figure 39.3.

Figure 39.3

It follows that AB = DC and BC = AD

Practice Problems

Problem 39.1Draw all lines of symmetry for the figure below.

Problem 39.2(a) Find the vertical line(s) of symmetry of the letters A, U, V, T, Y.(b) Find the horizontal line(s) of symmetry of the letters D, E, C, B.(c) Find the vertical and horizontal line(s) of symmetry of the letters H, I,O, X.

Problem 39.3For each figure, find all the lines of symmetry you can.

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Problem 39.4A regular polygon is a closed figure with all sides congruent. Find all thelines of symmetry for these regular polygons. Generalize a rule about thenumber of lines of symmetry for regular polygons.

Problem 39.5Find the number of rotations of the following geometric shape.

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Diagonals of QuadrilateralsA diagonal of a quadrilateral is a line segement formed by connecting non-adjacent vertices (i.e., not on the same side). Each quadrilateral discussedin Section 37 has two diagonals.

Example 39.2For each geoboard quadrilateral below draw in the two diagonals. Enter thedata requested below each figure. The meaning of each entry is given here.

Type: Which of the seven types of quadrilaterals best describes the fig-ure: square, rhombus, rectangle, parallelogram, kite, trapezoid or isoscelestrapezoid.Cong ∼=: Are the diagonals congruent? Yes or No.

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Perp ⊥: Are the diagonals perpendicular? Yes or No.Bisect: Do the diagonals bisect each other? Yes or No.Bisect ∠ : Do the diagonals bisect the corner angles? Yes or No.

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Solution.

CirclesSo far we have been studying plane figures with boundaries consisting of linesegments. Circles are plane figures with curved boundary. By definition, acircle is the set of all points in the plane that are equidistant from a fixed

187

point called the center of the circle. The distance from the center to a pointon the circle is called the radius. Any line segment crossing the center andwhose endpoints are on the circle is called a diameter of the circle. It isclear that the length of a diameter is twice the length of the radius of thecircle. See Figure 39.4.

Figure 39.4

A useful device for drawing circles is the compass as shown in Figure 39.5.

Figure 39.5

The steps for drawing a circle with a compass are as follows:

1. Insert a sharp pencil into the holder on the compass.2. Open the compass to the radius desired for your circle.3. Place the compass point on a piece of paper where you would like thecenter point of the circle to be.4. Place the point of the pencil on the paper.5. Rotate the compass to mark a circle on the paper with the pencil. Drawwithout lifting the point of the compass off the paper.

Finally, we conclude this section by looking at the symmetry properties ofcircles. Any diameter is a line of symmetry of a circle so that a circle hasinfinitely many lines of symmetry. Also, a circle has infinitely many rotationsymmetries, since every angle whose vertex is the center of the circle is anangle of rotation symmetry.

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Practice Problems

Problem 39.7Let ABCD be a parallelogram.(a) Prove that ∠A and ∠B are supplementary.(b) Prove that angles ∠A and ∠C are congruent.

Problem 39.8Using the figure below find the height of the trapezoid(i.e., the distancebetween the parallel sides) in terms of a and b.

Problem 39.9”The diagonals of a rectangle are congruent.” Why does this statement implythat the diagonals of a square must also be congruent?

Problem 39.10You learn the theorem that the diagonals of a parallelogram bisect each other.What other quadrilaterals must have this property?

Problem 39.11Fill in the blank to describe the following circle with center N. Circle N isthe set of in a plane that are from .

Problem 39.12A chord is a line segment with endpoints on a circle. True or false? Everychord is also a diameter of the circle.

Problem 39.13The diameter of a circle divides its interior into two congruent regions. Howcan someone use this property in dividing a circular pizza or pie in half?

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Problem 39.14If possible draw a triangle and a circle that intersect at exactly(a) one point (b) two points (c) three points (d) four points

Problem 39.15If possible draw two parallelograms that intersect at exactly(a) one point (b) two points (c) three points (d) four points

Problem 39.16A quadrilateral has two right angles. What can you deduce about the mea-sures of the other two angles?

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40 Regular Polygons

Convex and Concave ShapesA plane figure is said to be convex if every line segment drawn between anytwo points inside the figure lies entirely inside the figure. A figure that is notconvex is called a concave figure. Figure 40.1 shows a set of convex figures.

Figure 40.1

On the other hand, Figure 40.2 shows concave figures. To show that a figure isconcave, it is enough to find two points within the figure whose correspondingline segment is not completely inside the figure.

Figure 40.2

Regular PolygonsBy a closed curve we mean a curve that starts from one point and ends inthat same point. By a simple curve we mean a curve that does not crossitself. By a simple closed curve we mean a a curve in the plane that starts

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and ends in the same location without crossing itself.Several examples of curves are shown in Figure 40.3.

Figure 40.3

A polygon is a simple closed curve made up of line segments. A polygonwhose line segments are congruent and whose interior angles are all congruentis called a regular polygon. If a regular polygon consists of n sides thenwe will refer to it as regular n-gon.Figure 40.4 shows several tpyes of regular n-gons.

Figure 40.4

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Angles of interest in a regular n-gon are the following: A vertex angle (alsocalled an interior angle) is formed by two consecutive sides. A centralangle is formed by the segments connecting two consecutive vertices to thecenter of the regular n-gon. An exterior angle is formed by one side togetherwith the extension of an adjacent side as shown in Figure 40.5

Figure 40.5

Angles Measures in Regular PolygonsLet’s first find the measure of a central angle in a regular n-gon. Connectingthe center of the n-gon to the n vertices we create n congruent central an-gles. Since the sum of the measures of the n central angles is 360◦ then themeasure of each central angle is 360◦

n.

Next, we will find the measure of each interior angle of a regular n-gon. Wewill use the method of recognizing patterns for that purpose. Since the anglesare congruent then the measure of each is the sum of the angles divided by n.Hence, we need to find the sum of the interior angles. This can be achievedby dividing the n-gon into triangles and using the fact that the sum of thethree interior angles in a triangle is 180◦. The table below suggests a way forcomputing the measure of a vertex angle in a regular n-gon for n=3,4,5,6,7,8.

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So, in general, the measure of an interior angle of a regular n-gon is

(n− 2) · 180◦

n= 180◦ − 360◦

n

To measure the exterior angles in a regular n-gon, notice that the interiorangle and the corresponding adjacent exterior angle are supplementary. SeeFigure 40.6. Thus, the measure of each exterior angle is

180◦ − (n− 2) · 180◦

n=

360◦

n.

Figure 40.6

Example 40.1(a) Find the measure of each interior angle of a regular decagon (i.e., n =10).(b) Find the number of sides of a regular polygon, each of whose interiorangles has a measure of 175◦.

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Solution.(a) The measure of each angle is: (10−2)·180◦

10= 144◦.

(b) We are given that (n−2)·180◦

n= 175◦ or 180◦ − 360◦

n= 175◦. This implies

that 360◦

n= 180◦ − 175◦ = 5◦. Thus, n = 360◦

5◦= 72

Practice Problems

Problem 40.1List the numerical values of the shapes that are convex .

Problem 40.2Determine how many diagonals each of the following has:(a) 20-gon (b) 100-gon (c) n-gon

Problem 40.3In a regular polygon, the measure of each interior angle is 162◦. How manysides does the polygon have?

Problem 40.4Two sides of a regular octagon are extended as shown in the following figure.Find the measure of ∠1.

Problem 40.5Draw a quadrilateral that is not convex.

Problem 40.6What is the sum of the interior angle measures of a 40-gon?

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Problem 40.7A Canadian nickel has the shape of a regular dodecagon (12 sides). Howmany degrees are in each interior angle?

Problem 40.8Is a rectangle a regular polygon? Why or why not?

Problem 40.9Find the measures of the interior, exterior, and central angles of a 12-gon.

Problem 40.10Suppose that the measure of the interior angle of a regular polygon is 176◦.What is the measure of the central angle?

Problem 40.11The measure of the exterior angle of a regular polygon is 10◦. How manysides does this polygon have?

Problem 40.12The measure of the central angle of a regular polygon is 12◦. How many sidesdoes this polygon have?

Problem 40.13The sum of the measures of the interior angles of a regular polygon is 2880◦.How many sides does the polygon have?

Problem 40.14How many lines of symmetry does each of the following have?

(a) a regular pentagon(b) a regular octagon(c) a regular hexagon.

Problem 40.15How many rotational symmetry does a pentagon have?

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41 Three Dimensional Shapes

Space figures are the first shapes children perceive in their environment. Inelementary school, children learn about the most basic classes of space fig-ures such as prisms, pyramids, cylinders, cones and spheres. This sectionconcerns the definitions of these space figures and their properties.

Planes and Lines in SpaceAs a basis for studying space figures, first consider relationships among linesand planes in space. These relationships are helpful in analyzing and definingspace figures.Two planes in space are either parallel as in Figure 41.1(a) or intersect asin Figure 41.1(b).

Figure 41.1

The angle formed by two intersecting planes is called the dihedral angle.It is measured by measuring an angle whose sides line in the planes and areperpendicular to the line of intersection of the planes as shown in Figure41.2.

Figure 41.2

Some examples of dihedral angles and their measures are shown in Figure41.3.

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Figure 41.3

Two nonintersecting lines in space are parallel if they belong to a commonplane. Two nonintersecting lines that do not belong to the same plane arecalled skew lines. If a line does not intersect a plane then it is said to beparallel to the plane. A line is said to be perpendicular to a plane ata point A if every line in the plane through A intersects the line at a rightangle. Figures illustrating these terms are shown in Figure 41.4.

Figure 41.4

PolyhedraTo define a polyhedron, we need the terms ”simple closed surface” and”polygonal region.” By a simple closed surface we mean any surface with-out holes and that encloses a hollow region-its interior. An example is asphere, which is the set of all points at a constant distance from a singlepoint called the center. The set of all points on a simple closed surfacetogether with all interior points is called a solid. For example, the shell ofa hardboiled egg can be viewed as a simple closed surface whereas the shelltogether with the white and the yolk of the egg form a solid. A polygonalregion is a polygon together with its interior as shown in Figure 41.5.

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Figure 41.5

A polyhedron (polyhedra is the plural) is a simple closed surface boundedby polygonal regions. All of the surfaces are flat, not curved. The polygonalregions are called the faces. The sides of each face are called the edges. Thevertices of the polygons are also called vertices of the polyhedron. Figure41.6(a) and (b) are examples of polyhedra, but (c) and (d) are not.

Figure 41.6

Polyhedra are named according to the number of faces. For example, atetrahedron has four faces, a pentahedron has five faces, a hexahedronhas six faces, and so on.

199

Figure 41.7

Polyhedra can be classified into several types:• Prisms: A prism is a polyhedron, with two parallel faces called bases.The other faces are called lateral faces and are always parallelogram. If thelateral faces are rectangles then the prism is called a right prism. Otherwise,the prism is called an oblique prism. Figure 41.8 shows some types ofprisms.

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Figure 41.8

• Pyramids: A polyhedron is a pyramid if it has 3 or more triangular facessharing a common vertex, called the apex. The base of a pyramid may beany polygon. Pyramids are named according to the type of polygon formingthe base. Pyramids whose bases are regular polygons are called regularpyramids. If in addition, the lateral faces are isosceles triangles then thepyramids are called right regular pyramids. Otherwise, they are calledoblique regular pyramids. See Figure 41.9.

Figure 41.9

Regular PolyhedraA polyhedron is said to be regular if all its faces are identical regular polyg-onal regions and all dihedral angles have the same measures. Regular poly-hedra are also known as platonic solids. There are five regular polyhedraanalyzed in the table below and shown in Figure 41.10.

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Name Faces Edges VerticesTetrahedron 4 6 4Octahedron 8 12 6Hexahedron 6 12 8Icosahedron 20 30 12Dodecahedron 12 30 20

Curved Shapes: Cylinders, Cones, and SpheresSome everyday objects suggest space figures that are not polyhedra: cylinders(can of soup), cones (ice cream cone), and spheres (a basketball).• Cylinders: A cylinder is the space figure obtained by translating thepoints of a simple closed region in one plane to a parallel plane as shown inFigure 41.11(a). The simple closed regions are called the base of the cylinder.The remaining points constitute the lateral surface of the cylinder.The most common types of cylinders in our environment are right circularcylinders. The line joining the centers of the two circles is called the axis.If the axis is perpendicular to the circles then the cylinder is called a rightcircular cylinder (See Figure 41.11(b)). Otherwise it is called an obliquecircular cylinder. (See Figure 41.11(c))

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Figure 41.11

• Cones: Suppose we have a simple closed region in a plane and a point Pnot in the plane of the region. The union of line segments connecting P toeach point in the simple closed region is called a cone. See Figure 41.12(a).P is called the vertex. The simple closed region is called the base. Thepoints in the cone not in the base constitute the lateral surface. If the baseis a circle then the cone is called a circular cone. A circular cone where theline connecting the vertex to the center of the base is perpendicular to thebase is called a right circular cone. See Figure 41.12(b). Otherwise, thecone is called oblique circular cone. See Fibure 41.12(c).

Figure 41.12

• Spheres: We have already encountered this concept. Recall that a sphereis the collection of all points in the space that are the same distance from afixed point called center. Any line segment joining the center to a point onthe surface of the sphere is called a radius. Any line segment going through

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the center and connecting two points on the sphere is called a diameter.See Figure 41.13.

Figure 41.13

Practice Problems

Problem 41.1Determine the measures of all dihedral angles of a right prism whose basesare regular octagons.

Problem 41.2Consider the cube given in the figure below.

(a) How many planes are determined by the faces of the cube?(b) Which edges of the cube are parallel to edge AB?(c) Which edges of the cube are contained in lines that are skew to the linegoing through A and B?

Problem 41.3If a pyramid has an octagon for a base, how many lateral faces does it have?

Problem 41.4Determine the number of faces, vertices, and edges for a hexagonal pyramid.

Problem 41.5Sketch drawings to illustrate different possible intersections of a square pyra-mid and a plane.

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Problem 41.6Sketch each of the following.

(a) A plane and a cone that intersect in a circle.(b) A plane and a cylinder that intersect in a segment.(c) Two pyramids that intersect in a triangle.

Problem 41.7For the following figure, find the number of faces, edges, and vertices.

Problem 41.8For each of the following figures, draw all possible intersections of a planewith(a) Cube (b) Cylinder

Problem 41.9A diagonal of a prism is any segment determined by two vertices that do notlie in the same face. How many diagonals does a pentagonal prism have?

Problem 41.10For each of the following, draw a prism and a pyramid that have the givenregion as a base:

(a) Triangle(b) Pentagon(c) Regular hexagon

Problem 41.11If possible, sketch each of the following:(a) An oblique square prism

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(b) An oblique square pyramid(c) A noncircular right cone(d) A noncircular cone that is not right

Problem 41.12A simple relationship among the number of faces(F), the number of edges(E),and the number of vertices(V) of any convex polyhedron exists and is knownas Euler’s formula. Use the following table to find this relationship.


Problem 41.13What is the sum of the angles at the each vertex of a(a) Tetrahedron(b) Octahedron(c) Cube(d) Icosahedron(e) Dodecahedron

Problem 41.14Determine for each of the following the minimum number of faces possible:(a) Prism (b) Pyramid (c) Hexahedron

Problem 41.15True or false?(a) Through a given point not on plane P, there is exactly one line parallelto P.(b) Every set of four points is contained in one plane.(c) If a line is perpendicular to one of two parallel planes then it is perpen-dicular to the other.

Problem 41.16Tell whether each of the following suggests a polygon or a polygonal region.(a) A picture frame.(b) A page in this book.(c) A stop sign.

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Problem 41.17A certain prism has 20 vertices. How many faces and edges does it have?

Problem 41.18Why are there no such things as skew planes?

Problem 41.19A prism has a base with n sides.

(a) How many faces does it have?(b) How many vertices does it have?(c) How many edges does it have?

Problem 41.20A pyramid has a base with n sides.


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42 Systems of Measurements

So far only two goemetric figures have been measured: line segments, mea-sured by the distance between their endpoints; and angles, measured by thedegrees of rotation needed to turn one side to the other. In this and comingsections we will introduce more general notions of measurement of geometricfigures. All curves, not just segments will be given a length. Plane regionswill be measured by area and perimeter. Space figures will be measured bysurface area and volume.We begin by discussing the general process of measurement. The two prin-cipal systems of measurements are then described: the English System andthe Metric or International System.

The Measurement ProcessThe measurement process consists of the following steps:(1) Select an object and an attribute of the object to measure, such as itslength, area, volume, temperature, or weight.(2) Select an appropriate unit which to measure the attribute.(3) Determine the number of units needed to measure the attribute.

Measurements can be done using either nonstandard units or standard units.For example, the first nonstandard units of length were based on people’shands, feet, and arms. Since different people have different-sized hands, feet,and arms then these measurements are adequate for convenient but lack ac-curacy. This is why we need standard measuring units such as of those of themetric system. By using standard units, people around the globe communi-cate easily with one another about measurements. We consider two standardsystems of measurements:

The US Customary System(English System)The following are the common conversion equivalencies in the English Sys-tem.Length

1 in (inch) = 112

ft (foot)1 yd (yard) = 3 ft

1 rd (rod) = 16.5 ft1 furlong = 660 ft

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1 mi (mile) = 5280 ft

Area

1 square inch = 1144

square feet1 square yard = 9 square feet1 acre = 43,560 square feet

1 square mile = 27,878,400 square feet

Volume

1 cubic in = 11728

cubic feet1 cubic yard = 27 cubic feet

Capacity

1 tablespoon = 3 teaspoons1 ounce = 2 tablespoons

1 cup = 8 ounces1 pint = 2 cups

1 quart = 2 pints1 gallon = 4 quarts

1 barrel = 31.5 gallons

Weight

1 pound = 16 ounces1 ton = 2000 pounds

Time

1 minute = 60 seconds1 hour = 60 minutes

1 day = 24 hours1 week = 7 days

1 year ≈ 365.25 days

Temperature(Fahrenheit)

32◦F = freezing point of water212◦F = boiling point of water

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Dimensional AnalysisTo convert from one unit to another, the process knwon as dimensionalanalysis can be used. This process works with unit ratios( ratios equal to1). For example, 1 yd

3 ftand 5280 ft

1 miare units ratios. Therefore to convert 5.25

mi to yards, we have the following

5.25 mi = 5.25 mi× 5280 ft1 mi

× 1 yd3 ft

= 9240 yd

Example 42.1If a cheetah is clocked at 60 miles per hour, what is its speed in feet persecond?

Solution.

60mihr

= 60mihr× 5280 ft

1 mi× 1 hr

60 min× 1 min

60 sec= 88 ft

sec

Example 42.2Convert each of the following:(a) 219 ft = yd(b) 8432 yd = mi(c) 0.2 mi = ft(d) 64 in = yd

Solution.(a) 219 ft = 219 ft ×1 yd

3 ft= 73 yd.

(b) 8432 yd = 8432 yd × 3 ft1 yd

× 1 mi5280 ft

≈ 4.79 mi.

(c) 0.2 mi = 0.2 mi ×5280 ft1 mi

= 1056 ft

(d) 64 in = 64 in × 1 ft12 in

× 1 yd3 ft

≈ 1.78 yd

International SystemSix common prefixes used in the metric system are listed below.

Prefixes Multiple or fractionkilo 1000hecto 100deca 10deci 0.1centi 0.01milli 0.001

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The following are the common conversion equivalencies in the Metric System.

Length

Unit Abbreviation Number of Meterskilometer km 1,000hectometer hm 100dekameter dam 10meter m 1decimeter dm 0.1centimeter cm 0.01millimeter mm 0.001

Area

Unit Abbreviation Number ofsquare meters

square millimeter mm2 0.000001square centimeter cm2 0.0001square decimeter dm2 0.01Are a 100Hectare ha 10000square kilometer km2 1000000

Volume

Unit Abbreviation Number ofcubic meters

cubic meter m3 1cubic decimeter dm3 0.001cubic centimeter cm3 0.000001

Capacity

Unit Abbreviation Number of literskiloliter kL 1,000hectoliter hL 100dekaliter daL 10liter L 1deciliter dL 0.10centiliter cL 0.01milliliter(cm3) mL 0.001

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Mass

Unit Abbreviation Number of gramsmetric ton t 1,000,000kilogram kg 1,000hectogram hg 100dekagram dag 10gram g 1decigram dg 0.10centigram cg 0.01milligram mg 0.001

Temperature(Celsius)

0◦C = freezing point of water100◦C = boiling point of water

Example 42.3Convert the following measurements to the unit shown.

(a) 1495 mm = m(b) 29.4 cm = mm(c) 38741 m = km

Solution.(a) 1495 mm = 1495 mm× 1 m

1000 mm= 1.495 m

(b) 29.4 cm = 29.4 cm× 1 mm0.1 cm

= 294 mm(c) 38741 m = 38741 m× 1 km

1000 m= 38.741km

Conversion Between SystemsLength: 1 in = 2.54 cmArea: 1 in2 = 2.542 cm2

Volume: 1 in3 = 2.543 cm3

Capacity: 1 cm3 = 1 mL, 1 gal ≈ 3.79 LWeight: 1 oz = 29 g

Temperature: F = 95C + 32

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Practice Problems

Problem 42.1A small bottle of Perrier sparking water contains 33 cl. What is the volumein ml?

Problem 42.2Fill in the blanks.(a) 58728 g = kg(b) 632 mg = g(c) 0.23 kg = g

Problem 42.3Convert each of the following.(a) 100 in = yd(b) 400 yd = in(c) 300 ft = yd(d) 372 in = ft


Item m cm mmLength of a piece of paper 35Height of a woman 1.63Width of a filmstrip 35Length of a cigarette 100

Problem 42.5List the following in decreasing order: 8 cm, 5218 mm, 245 cm, 91 mm, 6 m,700 mm.

Problem 42.6Complete each of the following.(a) 10 mm = cm(b) 3 km = m(c) 35 m = cm(d) 647 mm = cm

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Problem 42.7Complete the following conversions.(a) 3 feet = inches(b) 2 miles = feet(c) 5 feet = yards

Problem 42.8Complete the following conversions.(a) 7 yards = feet(b) 9 inches = feet(c) 500 yards = miles

Problem 42.9Complete the following conversions.(a) 9.4 L = mL(b) 37 mg = g(c) 346 mL = L

Problem 42.10A nurse wants to give a patient 0.3 mg of a certain drug. The drug comes ina solution containing 0.5 mg per 2 mL. How many milliters should be used?

Problem 42.11A nurse wants to give a patient 3 gm of sulfisoxable. It comes in 500 mgtablets. How many tablets should be used?

Problem 42.12Complete the following conversions.(a) 3 gallons = quarts(b) 5 cups = pints(c) 7 pints = quarts(d) 12 cups = gallons

Problem 42.13True or false? Expain.(a) 1 mm is longer than 1 in.(b) 1 m is longer than 1 km.(c) 1 g is heavier than 1 lb.(d) 1 gallon is more than 1 L.

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Problem 42.14Derive a conversion formula for degrees Celsius to degrees Fahrenheit.

Problem 42.15A temperature of −10◦C is about(a) −20◦F (b) 10◦F (c) 40◦F (d) 70◦F

Problem 42.16Convert the following to the nearest degree(a) Moderate oven (350◦F ) to degrees Celsius.(b) 20◦C to degrees Fahrenheit.(c) −5◦C to degrees Fahrenheit.

Problem 42.17Complete the following conversions.(a) 1 cm2 = mm2

(b) 610 dam2 = hm2

(c) 564 m2 = km2

(d) 0.382 km2 = m2

Problem 42.18Suppose that a bullet train is traveling 200 mph. How many feet per secondsis it traveling?

Problem 42.19A pole vaulter vaulted 19ft41

2in. Find the height in meters.

Problem 42.20The area of a rectangular lot is 25375 ft2. What is the area of the lot inacres? Use the fact that 640 acres = 1 square mile.

Problem 42.21A vase holds 4286 grams of water. What is the capacity in liters? Recallthat the density of water is 1 g/cm3.

Problem 42.22By using dimensional analysis, make the following conversions.(a) 3.6 lb to oz(b) 55 mi/hr to ft/min(c) 35 mi/hr to in/sec(d) $575 per day to dollars per minute.

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Problem 42.23The density of a substance is the ratio of its mass to its volume. A chunk ofoak firewood weighs 2.85 kg and has a volume of 4100 cm3. Determine thedensity of oak in g/cm3, rounding to the nearest thousandth.

Problem 42.24The speed of sound is 1100 ft/sec at sea level. Express the speed of sound inmi/hr.

Problem 42.25What temperature is numerically the same in degrees Celsius and degreesFahrenheit?

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43 Perimeter and Area

Perimeters of figures are encountered in real life situations. For example, onemight want to know what length of fence will enclose a rectangular field. Inthis section we will study the perimeters of polygons and circles.By definition, the perimeter of a simple closed plane figure is the length ofits boundary. The perimeter is always measured in units of length, such asfeet or centimeters.The perimeter of a polygon is defined to be the sum of the lengths of itssides. Figure 43.1 exhibits the perimeters of some standard plane figures.We will use p to denote the perimeter of a plane figure.

Figure 43.1

The Circumference of a Circle:The perimeter of a circle is called its circumference. Using a tape measurearound the circle one can find the circumference. Also, one will notice thatthe ratio of the circumference of a circle to its diameter is the same for allcircles. This common ratio is denoted by π. Thus

Cd

= π or C = πd.

Since the diameter d is twice the radius r, we also have C = 2πr. It hasbeen shown that π is an irrational number with unending decimal expansion:π = 3.1415926 · · · .

Example 43.1A dining room table has 8 sides of equal length. If one side measures 15inches, what is the perimeter of the table?

217

Solution.If we denote the length of a side by a then the perimeter of the table isp = 8a. Since a = 15 in then p=15(8) = 120 in

Example 43.2The circumference of a circle is 6π cm. Find its radius.

Solution.Since C = 2πr and C = 6π then 2πr = 6π. Solving for r we find r = 6π

2π=

3 cm

Practice Problems

Problem 43.1An oval track is made up by erecting semicircles on each end of a 50-m by100-m rectangle as shown in the figure below.

What is the perimeter of the track?

Problem 43.2Find each of the following:(a) The circumference of a circle if the radius is 2 m.(b) The radius of a circle if the circumference is 15π m.

Problem 43.3Draw a triangle ABC. Measure the length of each side. For each of thefollowing, tell which is greater?(a) AB + BC or AC(b) BC + CA or AB(c) AB + CA or BC

Problem 43.4Can the following be the lengths of the sides of a triangle? Why or Why not?(a) 23 cm, 50 cm, 60 cm(b) 10 cm, 40 cm, 50 cm(c) 410 mm, 260 mm, 14 cm

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Problem 43.5Find the circumference of a circle with diameter 6π cm.

Problem 43.6What happens to the circumference of a circle if the radius is doubled?

Problem 43.7Find the length of the side of a square that has the same perimeter as arectangle that is 66 cm by 32 cm.

Problem 43.8A bicycle wheel has a diameter of 26 in. How far a rider travel in one fullrevolution of the tire? Use 3.14 for π.

Problem 43.9A car has wheels with radii of 40 cm. How many revolutions per minutemust a wheel turn so that the car travels 50 km/h?

Problem 43.10A lot is 21 ft by 30 ft. To support a fence, an architect wants an uprightpost at each corner and an upright post every 3 ft in between. How many ofthese posts are needed?

AreaThe number of units required to cover a region in the plane is known as itsarea. Usually squares are used to define a unit square. An area of ”10square units” means that 10 unit squares are needed to cover a flat surface.Below, we will find the area of some quadrilaterals and the area of a circle.

Area of a rectangle:A 3-cm by 5-cm rectangle can be covered by 15 unit squares when the unitsquare is 1 cm2, as shown in Figure 43.2(a). Similarly a 2.5-cm by 3.5 cmrectangle can be covered by six whole units, five half-unit squares, and onequarter unit square, giving a total area of 8.75 cm2 as shown in Figure 43.2(b).This is also the product of the width and the length since 2.5×3.5 = 8.75 cm2.It follows that a rectangle of length L and width W has an area A given bythe formula

A = LW.

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Figure 43.2

Example 43.3The area of a rectangle of length L and width W is equal to its perimeter.Find the relationship between L and W.

Solution.Since A = LW and p = 2(L + W ) then LW = 2(L + W ). That is, LW =2L + 2W. Subtracting 2L from both sides we obtain L(W − 2) = 2W. Thus,L = 2W

W−2

Area of a square:Since a square is a rectangle with length equals width then the area of asquare with side length equals to s is given by the formula

A = s2.

Area of a triangle:First, we consider the case of a right triangle as shown in Figure 43.3(a).Construct rectangle ABDC where ∆DCB is a copy of ∆ABC as shown inFigure 43.3(b). The area of the rectangle ABDC is bh, and the area of ∆ABCis one-half the area of the rectangle. Hence, the area of ∆ABC = 1

2bh.

Now, suppose we have an arbitrary triangle as shown in Figure 43.3(c). Then

Area of ∆ABC = area of ∆ADC + area of ∆CDB

That is,A = 1

2h · AD + 1

2h ·DB

= h2(AD + DB)

= h·AB2

220

Figure 43.3

Area of a parallelogram:Consider a parallelogram with a pair of opposite sides b units long, and thesesides are h units apart as shown in Figure 43.4(a). The side with lengthb is called the base of the parallelogram and h is the height or altitude.Removing and replacing a right triangle as shown in Figure 43.4(b) results ina rectangle of length b and width h with the same area as the parallelogram.This means that the area of the parallelogram is given by the formula

A = bh.

Figure 43.4

Example 43.4Suppose a fifth grader has not yet learned the area formula for a parallelo-gram. How could you develop the formula from the geoboard diagram?

Solution.The geoboard shows that the length of the base is 3 and the height is 2.Using unit parallelogram we see that 6 complete unit parallelograms coverthe inside of the large parallelogram. This shows that the area is 2 · 3 = 6

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Area of a Rhombus:Consider the rhombus given in Figure 43.5. Since the diagonals bisect eachother at right angles then we have

TQ = TS = QS2

and TP = TR = PR2

But

Area of rhombus = area of triangle PQR + area of triangle PRS

Thus,A = 1

2TQ · PR + 1

2TS · PR

= PR2

(TQ + TS)

= PR·QS2

Figure 43.5

Area of a kite:Consider the kite shown in Figure 43.6. Then

Area of kite = area of triangle DAB + area of triangle DCB

Thus,A = 1

2TA ·BD + 1

2TC ·BD

= BD2

(TA + TC)

= BD·AC2

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Figure 43.6

Area of a trapezoid:Consider the trapezoid shown in Figure 43.7(a). Consider two identical trape-zoids, and ”turn” one around and ”paste” it to the other along one side aspictured in Figure 43.7(b).

Figure 43.7

The figure formed is a parallelogram having an area of h(a + b), which istwice the area of one of the trapezoids. Thus, the area of a trapezoid withheight h and parallel sides a and b is given by the formula

A =h(a + b)

2.

Area of a circle:We will try to discover the formula of the area of a circle, say of radius 7 cm,by executing the following steps:Step 1: Using a compass, draw a circle of radius 7 cm. Then mark thecircle’s centre and draw its radius. See Figure 43.8(a).Step 2: Place the centre of the protractor at the centre of the circle and thezero line along the radius. Then mark every 30◦ around the circle.Step 3: Using a ruler and a pencil, draw lines joining each 30◦ mark to the

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centre of the circle to form 6 diameters. The diagram thus obtained will have12 circular sectors as shown in Figure 43.8(c).Step 4: Colour the parts as shown in Figure 43.8(d).Step 5: Cut out the circle and then cut along the diameters so that all parts(i.e. sectors) are separated. See Figure 43.8(e).Step 6: Using a ruler, measure the base and the height of the approximateparallelogram obtained in Step 5. See Figure 43.8(f).

Figure 43.8

Thus, the base is approximately one-half of the circumference of the circle,that is b = 7π. The height of the parallelogram is just the radius. So h = 7.Hence, the area of the circle is about 49π = π72.In general, the area of a circle of radius r is given by the formula

A = πr2.

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The Pythagorean FormulaGiven a right triangle with legs of lengths a and b and hypotenuse of lengthc as shown in Figure 43.9(a). The Pythagorean formula states that the sumof the squares of the lengths of the legs of a right triangle is equal to thesquare of the length of the hypotenuse:

c2 = a2 + b2.

The justification of this is as follows: Start with four copies of the sametriangle. Three of these have been rotated 90◦, 180◦, and 270◦, respectivelyas shown in Figure 43.9(b). Each has area ab/2. Let’s put them together sothat they form a square with side c as shown in Figure 43.9(c).The square has a square hole with the side (a − b). Summing up its area(a− b)2 and the area of the four triangles 4

(ab2

)= 2ab we get

c2 = (a− b)2 + 2ab = a2 − 2ab + b2 + 2ab = a2 + b2.

Figure 43.9

Practice Problems

Problem 43.11Convert each of the following:

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(a) 1 cm2 = mm2

(b) 124,000,000 m2 = km2

Problem 43.12Find the cost of carpeting a 6.5 m × 4.5 m rectangular room if one metersquare of carpet costs $ 13.85.

Problem 43.13A rectangular plot of land is to be seeded with grass. If the plot is 22 m ×28 m and 1-kg bag of seed is needed for 85 m2 of land, how many bags ofseed are needed?

Problem 43.14Find the area of the following octagon.

Problem 43.15(a) If a circle has a circumference of 8π cm, what is its area?(b) If a circle of radius r and a square with a side of length s have equalareas, express r in terms of s.

Problem 43.16A circular flower bed is 6 m in diameter and has a circular sidewalk aroundit 1 m wide. Find the area of the sidewalk.

Problem 43.17(a) If the area of a square is 144 cm2, what is its perimeter?(b) If the perimeter of a square is 32 cm, what is its area?

Problem 43.18Find the area of each of the following shaded parts. Assume all arcs arecircular. The unit is cm.

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Problem 43.19For the drawing below find the value of x.

Problem 43.20The size of a rectangular television screen is given as the length of the diag-onal of the screen. If the length of the screen is 24 cm and the width is 18cm, what is the diagonal length?

Problem 43.21If the hypotenuse of a right triangle is 30 cm long and one leg is twice aslong as the other, how long are the legs of the triangle?

Problem 43.22A 15-ft ladder is leaning against a wall. The base of the ladder is 3 ft fromthe wall. How high above the ground is the top of the ladder?

Problem 43.23Find the value of x in the following figure.

Problem 43.24Find x and y in the following figure.

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Problem 43.25Complete the following table which concerns circles:

Radius Diameter Circumference Area5 cm

24 cm17π m2

20πcm

Problem 43.26Suppose the housing authority has valued the house shown here at $220 perft2. Find the assessed value.

Problem 43.27Two adjacent lots are for sale. Lot A cost $ 20,000 and lot B costs $27,000.Which lot has the lower cost per square meter?

Problem 43.28If the radius of a circle increases by 30% then by what percent does the areaof the circle increase?

Problem 43.29Find the area of the shaded region.

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44 Surface Area

The surface area of a space figure is the total area of all the faces of thefigure. In this section, we discuss the surface areas of some of the space fig-ures introduced in Section 41.

Right PrismsLet’s find the surface area of the right prism given in Figure 44.1.

Figure 44.1

As you can see, we can dissamble the prism into six rectangles. The totalarea of these rectangles which is the surface area of the box is

SA = 2lw + 2h(w + l).

If we let A denote the area of the base and P the perimeter of the base thenA = lw and P = 2(l + w). Thus,

SA = 2A + Ph.

This formula is valid for any right prism with height h, area of base A, andperimeter of base P.

Right CylindersTo find the surface area of a right cylinder with height h and radius of base rwe dissamble the cylinder into a rectangle and two circles as shown in Figure44.2

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Figure 44.2

Thus, the surface area is the sum of the area of the rectangle together withthe areas of the two circles. That is,

SA = 2πr2 + 2πr · h = 2πr(r + h).

Example 44.1A thin can has a height of 15 cm and a base radius of 7 cm. It is filledwith water. Find the total surface area in contact with the water. (Takeπ = 22/7)

Solution.

SA = 2π(7)(7 + 15) = 2 · 22

7· 7 · 22 = 968 cm2

Example 44.2The lateral surface area of a solid cylinder is 880 cm2 and its height is 10 cm.Find the circumference and area of the base of the cylinder. Take π = 22

7.

Solution.Since the lateral surface of a cylinder is height × perimeter of the base thenP = 880

10= 88 cm. Now, the radius of the circle is r = 88

2π= 44 · 7

22= 14 cm.

Finally, the area of the base is A = π142 = 227· 196 = 616 cm2

PyramidsThe surface area of a pyramid is obtained by summing the areas of the faces.

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We illustrate this for the square pyramid shown in Figure 44.3

Figure 44.3

The area of the base is b2 whereas the total area of the lateral faces is 4× bl2

=2bl. Thus,

SA = b2 + 2bl.

If we let A denote the area of the base and P the perimeter of the base then

SA = A +1

2Pl.

This formula is valid for any right pyramid. We call l the Slant height.

Example 44.3A right pyramid with slant height of 12 cm stands on a square base of sides10 cm. Calculate the total surface area.

Solution.

SA = 102 + 2(10)(12) = 100 + 240 = 340 cm2

Right ConesFirst, we define the slant height of a right circular cone to be the lengthof a straight line drawn from any point on the perimeter of the base to theapex. See Figure 44.4. If the radius of the base is r and the altitude of thecone is h, then by applying the Pythagorean formula we find that the lengthof the slant height is given by

l =√

r2 + h2

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Figure 44.4

To find the surface area of the cone, we cut it along a slant height and openit out flat to obtain a circular sector shown in Figure 44.5

Figure 44.5

Thus, the lateral surface area is the area of the circular sector. But

Area of sector ABCArea of circle with center C

= Arc length of ABCircumference of circle with center at C

Area of sector ABCπl2

= 2πr2πl

Area of sector ABC = πrl

But

Total surface area = Area of base + lateral surface area.

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HenceSA = πr2 + πrl

= πr(r + l)

= πr(r +√

r2 + h2)

SpheresArchimedes discovered that a cylinder that circumscribes a sphere, as shownin Figure 44.6, has a lateral surface area equal to the surface area, SA, of thesphere.

Figure 44.6

Thus,SA = 2πrh

= 2πr × 2r= 4πr2

Example 44.4A solid sphere has a radius of 3 m. Calculate its surface area. Round youranswer to a whole number. (Take π = 22/7)

Solution.SA = 4πr2

= 4× 227× 32

≈ 113 m2

Example 44.5Find the radius of a sphere with a surface area of 64π m2.

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Solution.SA = 4πr2

64π = 4πr2

16 = r2

r = 4 cm

Practice Problems

Problem 44.1A small can of frozen orange juice is about 9.5 cm tall and has a diameterof about 5.5 cm. The circular ends are metal and the rest of the can iscardboard. How much metal and how much cardboard are needed to makea juice can?

Problem 44.2A pyramid has a square base 10 cm on a side. The edges that meet the apexhave length 13 cm. Find the slant height of the pyramid, and then calculatethe total surface area of the pyramid.

Problem 44.3An ice cream cone has a diameter of 2.5 in and a slant height of 6 in. Whatis the lateral surface area of the cone?

Problem 44.4The diameter of Jupiter is about 11 times larger than the diameter of theplanet Earth. How many times greater is the surface area of Jupiter?

Problem 44.5Find the surface area of each of the right prisms below.

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Problem 44.6The Great Pyramid of Cheops is a right square pyramid with a height of 148m and a square base with perimeter of 930 m. The slant height is 188 m. Thebasic shape of the Transamerica Building in San Fransisco is a right squarepyramid that has a height of 260 m and a square base with a perimeter of140 m. The altitude of slant height is 261 m. How do the lateral surfaceareas of the two structures compare?

Problem 44.7Find the surface area of the following cone.

Problem 44.8The Earth has a spherical shape of radius 6370 km. What is its surface area?

Problem 44.9Suppose one right circular cylinder has radius 2 m and height 6 m and an-other has radius 6 m and height 2 m.

(a) Which cylinder has the greater lateral surface area?(b) Which cylinder has the greater total surface area?

Problem 44.10The base of a right pyramid is a regular hexagon with sides of length 12 m.The height of the pyramid is 9 m. Find the total surface area of the pyramid.

Problem 44.11A square piece of paper 10 cm on a side is rolled to form the lateral surfacearea of a right circular cylinder and then a top and bottom are added. Whatis the surface area of the cylinder?

Problem 44.12The top of a rectangular box has an area of 88 cm2. The sides have areas 32cm2 and 44 cm2. What are the dimensions of the box?

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Problem 44.13What happens to the surface area of a sphere if the radius is tripled?

Problem 44.14Find the surface area of a square pyramid if the area of the base is 100 cm2

and the height is 20 cm.

Problem 44.15Each region in the following figure revolves about the horizontal axis. Foreach case, sketch the three dimensional figure obtained and find its surfacearea.

Problem 44.16The total surface area of a cube is 10,648 cm2. Find the length of a diagonalthat is not a diagonal of a face.

Problem 44.17If the length, width, and height of a rectangular prism is tripled, how doesthe surface area change?

Problem 44.18Find the surface area of the following figure.

Problem 44.19A room measures 4 meters by 7 meters and the ceiling is 3 meters high. Aliter of paint covers 40 square meters. How many liters of paint will it taketo paint all but the floor of the room?

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Problem 44.20Given a sphere with diameter 10, find the surface area of the smallest cylindercontaining the sphere.

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45 Volume

Surface area measures the area of the two-dimensional boundary of a three-dimensional figure; it is the area of the outside surface of a solid. Volume,on the other hand, is a measure of the space a figure occupies, or the space”inside” the three-dimensional figure. ”How much cardboard does it taketo make the box?” is a question about surface area; ”How much cereal canbe put into the box?” is a question about volume. Surface area is mea-sured in squares of length units, such as square feet (ft2), square centimeters(cm2) or square inches (in2). However, volume is measured in cubes of lengthunits, such as cubic feet (ft3), cubic centimeters (cm3) or cubic inches (in3).

Rectangular PrismsConsider a rectangular box of length 4 cm , width 2 cm, and height 3 cm.Then this box occupy the same space as 24 boxes each of length 1 cm, width1 cm, and height 1 cm. Note that 2 × 4 × 3 = 24. Thus, the volume of thebox is V = 24 cm3. This suggests that the volume of a rectangular prismwith length l, width w, and height h is given by the formula

V = lwh.

Since lw is the area B of the base of the prism then we can say that

Volume of rectangular prism = area of base × height

This formula is valid for any prism whose base area is B and height is h

Remark 45.1Note that a cube is a rectangular prism with l = w = h = s so that itsvolume is V = s3

Example 45.1Find the volume of the prism given below.

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Solution.First we have to find the area of the triangle that forms the base of thisprism. The area of a triangle is one-half the base of the triangle times theheight of the triangle. The triangle has a base of 14 inches and a height of12 inches. If we substitute these values into the formula, we get 84 squareinches for the area of the triangle.

B =1

2× 14× 12 = 84 in2

Since h = 20 in then the total volume is

V = 84× 20 = 1680 in3

Right PyramidsConsider a pyramid with rectangular base of area B and height h. Also,consider a rectangular box of area of base B and height h. Fill the pyramidwith rice and pour the content into the box. Repeat this process three times.You will notice that the cube is completely full with rice. This shows thatthe volume of the pyramid is one-third that of the box. That is,

V =1

3Bh.

This formula is true for any right pyramid.

Example 45.2Find the volume of a rectangular-based pyramid whose base is 8 cm by 6 cmand height is 5 cm.

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Solution.The volume is

V =1

3× 8× 6× 5 = 80 cm3

Right CylindersSince we know how to calculate the volume of a prism, we use a regular prismto approach a cylinder with base radius r and height h: a regular prism whosebase is a quadrilateral, a pentagon, a hexagon, a heptagon, an octagon, ora many-sided regular polygon. As the number of the sides increases, theperimeter of the regular polygon approaches the circumference of the circle,and the area of the base polygon approaches the base area of the cylinder.As a result, the volume of the regular prism approaches the volume of thecylinder as shown in Figure 45.2

Figure 45.2

Eventually, the regular prism and the cylinder fit together perfectly when thenumber of the sides is large enough. Since the volume of the regular prismis the product of its base area and its height, and since its base area andthe cylinder base fit together perfectly, the volume of the cylinder is also theproduct of its base area and its height. Since the base area is πr2 then thevolume of the cylinder is:

V = πr2h.

Example 45.3A steel pipe is 35.0 cm long and has an inside radius of 8.0 cm and an outsideradius of 10.0 cm. How much steel is needed to build the pipe? (The density

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of steel is about 7.8 g/cm3)

Solution.Given:

Inner radius r1 = 8.0 cmOuter radius r2 = 10.0 cm

Steel pipe length h = 35.0 cm

The volume of the pipe is the difference of two cylinders: one with the radiusof 10.0 cm and the other one with the radius of 8.0 cm. That is,

V = πr22h− πr2

1h= 3.14× 10.02× 35.0− 3.14× 8.02× 35.0= 3956.4 cm3

The weight of the pipe:

w = V olume× density = 30859.9 9 ≈ 30.86 kg

Right Circular ConesConsider a cone with radius r and height h. Take a cylinder of radius r andheight h. Fill the cone with rice til its top and pour the content into thecylinder. Repeat this process three times. After the third time you willnotice that the cylinder is completely full. This shows that the volume of thecone is one third the volume of the cylinder. But the volume of the cylinderis πr2h. Hence, the volume of the cone is

V =1

3πr2h

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SpheresWe find the volume of a sphere as follows: Take half a sphere of radius r andfill it completely with rice. Empty the rice into a right cylinder of radius rand height r. You will notice that the rice fill up 2

3of the cylinder. This shows

that the volume of the hemisphere is equal to 2/3 the volume of the cylinder.But the volume of the cylinder is πr3. Hence, the volume of a hemisphere is23πr3. The volume of the sphere is twice the volume of a hemisphere and is

given by the formula

V =4

3πr3.

Practice Problems

Problem 45.1Find the volume of each figure below.


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Problem 45.3Maggie is planning to build a new one-story house with floor area of 2000ft2. She is thinking of putting in a 9-ft ceiling. If she does this, how manycubic feet of space will she have to heat or cool?

Problem 45.4Two cubes have sides lengths 4 cm and 6 cm, respectively. What is the ratioof their volumes?

Problem 45.5What happens to the volume of a sphere if its radius is doubled?

Problem 45.6An olympic-sized pool in the shape of a right rectangular prism is 50 m ×25 m. If it is 2 m deep throughout, how many liters of water does it hold?Recall that 1 m3 = 1000 L.

Problem 45.7A standard straw is 25 cm long and 4 mm in diameter. How much liquid canbe held in the straw at one time?

Problem 45.8The pyramid of Khufu is 147 m high and its square base is 231 m on eachside. What is the volume of the pyramid?

Problem 45.9A square right regular pyramid is formed by cutting, folding, and gluing thefollowing pattern.

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(a) What is the slant height of the pyramid?(b) What is the lateral surface area of the pyramid?(c) Use the Pythagorean formula to find the height of the pyramid.(d) What is the volume of the pyramid?

Problem 45.10A cube 10 cm on a side holds 1 liter. How many liters does a cube 20 cm ona side hold?

Problem 45.11A right circular cone has height r and a circular base of radius 2r. Comparethe volume of the cone to that of a sphere of radius r.

Problem 45.12A store sell two types of freezers. Freezer A costs $350 and measures 2 ft by2 ft by 4.5 ft. Freezer B cots $480 and measures 3 ft by 3 ft by 3.5 ft. Whichfreezer is the better buy?

Problem 45.13Write a sentence that tells the difference between the surface area and volumeof a prism.

Problem 45.14A cylindrical water tank has a radius of 6.0 m. About how high must befilled to hold 400.0 m3?

Problem 45.15Roll an 8.5 by 11 in sheet of paper into a cylindrical tube. What is thediameter?

Problem 45.16A cylindrical pipe has an inner radius r, an outer radius R, and length l.Find its volume.

Problem 45.17A basketball has a diameter of 10 in. What is its volume?

Problem 45.18A standard tennis can is a cylinder that holds three tennis ball.

(a) Which is greater the circumference of the can or its height?(b) Find the volume of the can?

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Problem 45.19A cylindrical aquarium has a circular base with diameter 2 ft and height 3ft. How much water does it hold, in cubic feet?

Problem 45.20The circumference of a beach ball is 73 inches. How many cubic inches of airdoes the ball hold? Round your answer to the nearest cubic inch.

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46 Congruence of Triangles

Two triangles are congruent if one can be moved on top of the other, so thatedges and vertices coincide. The corresponding sides have the same lengths,and corresponding angles are congruent. That is, two triangles ∆ABC and∆A′B′C ′ are congruent if m(∠A) = m(∠A′), m(∠B) = m(∠B′), m(∠C) =m(∠C ′), AB = A′B′, AC = A′C ′, and BC = B′C ′. We write ∆ABC ∼=∆A′B′C ′.

Example 46.1Suppose that ∆ABC ∼= ∆A′B′C ′, AB = A′B′, AB = 2x + 10, and A′B′ =4x− 20. Find x.

Solution.Since AB = A′B′ then

4x− 20 = 2x + 104x− 2x = 10 + 20

2x = 30x = 15

Remark 46.1It is very important to maintain the vertices in the proper order. Not doingso is a common mistake.

Simpler conditions can be applied to verify that two triangles are congruent.The first one involves two sides and the included angle. We adopt this resultas an axiom, that is we accept this result as true by assumption, not by aproof as in the case of a theorem.

Axiom(Side-Angle-Side)If two triangles have two sides and the included angles equal, respectively,then the triangles are congruent.

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Example 46.2In ∆ABC, if AB = AC then m(∠B) = m(∠C).

Solution.Consider the correspondence of vertices A ↔ A, B ↔ C, and C ↔ B. Underthis correspondence, two sides and the included angle of ∆ABC are congru-ent respectively to the corresponding sides and included angle of ∆ACB.Hence, by SAS the triangles are congruent. Therefore, the correspondingangles are congruent and m(∠B) = m(∠C)

The second congruence property that we consider involves two angles andthe side included.

Theorem 46.1 (Angle-Side-Angle)If two angles and the included side of a triangle are congruent, respectively,to two angles and the included side of another triangle, then the two trianglesare congruent.

Proof.We start with two triangles ∆ABC and ∆A′B′C ′, where m(∠A) = m(∠A′), AB =A′B′, and m(∠B) = m(∠B′). We will show BC = B′C ′ and then apply SAStheorem above. (The same sort of argument shows AC = A′C ′.) In relat-ing BC and B’C’, there are three possibilities: BC = B’C’, BC < B′C ′, orBC > B′C ′. If the first case holds, we are done. Now suppose BC < B′C ′,then we can find a point G between B’ and C’ so that BC = B′G, andthen ∆ABC ∼= ∆A′B′G by SAS. In particular, this implies m(∠C ′A′B′) =m(∠CAB) = m(∠GA′B′). However, the segement A′G lies inside the angle∠C ′A′B′, so m(∠GA′B′) < m(∠C ′A′B′), which contradicts m(∠GA′B′) =m(∠C ′A′B′). So we cannot have BC < B′C ′. A similar argument arrives at acontradiction if BC > B′C ′, and so the only possibility is that BC = B′C ′

Example 46.3In Figure 46.1, we have AC = CD. Show that ∆ABC ∼= ∆DEC.

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Figure 46.1

Solution.Note that AD is a transversal line crossing two parallel lines. Then m(∠ACB) =m(∠DCE) (vertical angles) and m(∠CAB) = m(∠CDE) (alternate interiorangles). Moreover, AC = CD so by ASA the two triangles are congruent

The third property of congruence involves the three sides of the triangles.

Theorem 46.2 (Side-Side-Side)If a triangle has all three sides congruent to the corresponding sides of asecond triangle, then the two triangles are congruent.

Proof.Let B′′ be the point such that m(∠ACB′′) = m(∠A′C ′B′) and m(∠CAB′′) =m(∠C ′A′B′) and B′′ 6= B. See Figure 46.2

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Figure 46.2

This implies by angle-side-angle congruence that ∆AB′′C ∼= ∆A′B′C ′. Alsotriangles ∆ABB′′ and ∆CBB′′ are two isosceles triangles. Thus, the base an-gles are congruent. This implies m(∠ABB′′) = m(∠AB′′B) and m(∠CBB′′) =m(∠CB′′B). Thus, m(∠ABC) = m(∠AB′′C). Hence, by SAS, ∆ABC ∼=∆AB′′C. Therefore triangles ABC and A′B′C ′ are congruent

Example 46.4In Figure 46.3, we are given that AC = CD, AB = BD. Show that ∆ABC ∼=∆DBC.

Figure 46.3

Solution.Since AC=CD, AB=BD, and BC = CB then by SSS we have ∆ABC ∼=∆DBC

Practice Problems

Problem 46.1Suppose ∆JKL ∼= ∆ABC, where ∆ABC is shown below.

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Find the following(a) KL (b) LJ (c) m(∠L) (d) m(∠J)

Problem 46.2Using congruence of triangles show that equilateral triangles are equiangular.

Problem 46.3Let the diagonals of a parallelogram ABCD intersect at a point M.(a) Show that ∆ABM ∼= ∆CDM.(b) Use part (a) to explain why M is the midpoint of both diagonals of theparallelogram.

Problem 46.4In the figure below, AB = AE and AC=AD.

(a) Show that m(∠B) = m(∠E)(b) Show that m(∠ACD) = m(∠ADC)(c) Show that ∆ABC ∼= ∆AED(d) Show that BC = DE.

Problem 46.5Consider the following figure.

251

(a) Find AC(b) Find m(∠H), m(∠A), and m(∠C).

Problem 46.6Show that if ∆ABC ∼= ∆A′B′C ′ and ∆A′B′C ′ ∼= ∆A′′B′′C ′′ then ∆ABC ∼=∆A′′B′′C ′′.

Problem 46.7In the figure below, given that AB=BC=BD. Find m(∠ADC).

Problem 46.8In the figure below given that AB = AC. Find m(∠A).

Problem 46.9Find all missing angle measures in each figure.

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Problem 46.10An eighth grader says that AB=AC=AD , as shown in the figure below, thenm(∠B) = m(∠C) = m(∠D). Is this right? If not, what would you tell thechild?

Problem 46.11What type of figure is formed by joining the midpoints of a rectangle?

Problem 46.12If two triangles are congruent what can be said about their perimeters? ar-eas?

Problem 46.13In a pair of right triangles, suppose two legs of one are congruent to respec-tively to two legs of the other. Explain whether the triangles are congruentand why.

Problem 46.14A rural homeowner had his television antenna held in place by three guywires, as shown in the following figure. If the distance to each of the stakesfrom the base of the antenna are the same, what is true about the lengths ofthe wires? Why?

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Problem 46.15For each of the following, determine whether the given conditions are suffi-cient to prove that ∆PQR ∼= ∆MNO. Justify your answer.(a) PQ = MN, PR = MO, m(∠P ) = m(∠M)(b) PQ = MN, PR = MO, QR = NO(c) PQ = MN, PR = MO, m(∠Q) = m(∠N)

Problem 46.16Given that ∆RST ∼= ∆JLK, complete the following statements.(a) ∆TRS ∼= ∆(b) ∆SRT ∼= ∆(c) ∆TSR ∼= ∆(d) ∆JKL ∼= ∆

Problem 46.17You are given ∆RST and ∆XY Z with m(∠S) = m(∠Y ).(a) To show ∆RST ∼= ∆XY Z by the SAS congruence property, what morewould you need to know?(b) To show that ∆RST ∼= ∆XY Z by the ASA congruence property, whatmore would you need to know?

Problem 46.18You are given ∆ABC and ∆GHI with AB = GH. To show that ∆ABC ∼=∆GHI by the SSS congruence property, what more would you need to know?

Problem 46.19Suppose that ABCD is a kite with AB = AD and BC = DC. Show that thediagonal AC divides the kite into two congruent triangles.

Problem 46.20(a) Show that the diagonal of a parallelogram divides it into two congruenttriangles.(b) Use part (a) to show that the opposite sides of a parallelogram are con-gruent.(c) Use part (a) to show that the opposite angles of a parallelogram arecongruent.

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47 Similar Triangles

An overhead projector forms an image on the screen which has the sameshape as the image on the transparency but with the size altered. Two fig-ures that have the same shape but not necessarily the same size are calledsimilar. In the case of triangles, this means that the two triangles will havethe same angles and their sides will be in the same proportion (for example,the sides of one triangle might all be 3 times the length of the sides of theother). More formally, we have the following definition:Two triangles ∆ABC and ∆DEF are similar, written ∆ABC ∼ ∆DEF ifand only if(i) m(∠A) = m(∠D), m(∠B) = m(∠E), and m(∠C) = m(∠F )

(ii) ABDE

= ACDF

= BCEF

.

The common ratio in (ii) is called the scaled-factor. An example of twosimilar triangles is shown in Figure 47.1

Figure 47.1

Example 47.1Suppose in Figure 47.1 that EF = 6 cm, BC = 2cm and AB = 3cm. Whatis the value of DE?

Solution.Since DE = 3AB and AB = 3 then DE = 9 cm

The proofs of various properties of similar triangles depend upon certainproperties of parallel lines. Mainly, we need the following theorem which westate without proof.

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Theorem 47.1Let PQ be a line intersecting an angle ∠BAD. Then AP

AB= AQ

ADif and only if

the lines PQ and BD are parallel.

When we attempted to prove two triangles to be congruent we had a few testsSSS, SAS, ASA. In a similar way we have a few tests to help us determinewhether two triangles are similar.If the measures of two angles of a triangle are given, then the measure ofthe third angle is known automatically. Thus, the shape of the triangle iscompletely determined. Since similar triangles have the same shape, we havethe following similarity condition.

Theorem 47.2 (AA Test)If two angles of one triangle are congruent with the corresponding two anglesof another triangle, then the two triangles are similar.

Proof.Let ABC ↔ DEF denote a correspondence of ∆ABC and ∆DEF inwhich m(∠A) = m(∠D) and m(∠B) = m(∠E). Then m(∠A) + m(∠B) =m(∠D)+m(∠E). But 180◦−m(∠C) = m(∠A)+m(∠B) and 180◦−m(∠F ) =m(∠D) + m(∠E). Hence, m(∠C) = m(∠F ).Now, let F ′ be a point on DF and E ′ be a point on DE such that DF ′ = ACand DE ′ = AB. Then by SAS test for congruent triangles we have ∆ABC ∼=DE ′F ′. Hence, m(∠C) = m(∠DF ′E ′) and m(∠DF ′E ′) = m(∠DFE). Thislast equality implies that FE||F ′E ′ by the Corresponding Angles Theorem.

256

Hence, Theorem 47.1 implies

DF ′

DF=

DE ′

DE.

Since DF ′ = AC and DE ′ = AB then we have

AC

DF=

AB

DE.

Now, let D′ be a point on FE such that FD′ = CB. Then by repeating theabove argument we find

AC

DF=

BC

FE.

It follows thatAC

DF=

BC

FE=

AB

DE.

This shows ∆ABC ∼ ∆DEF

Example 47.2Consider Figure 47.2.

Figure 47.2

Show that BDCE

= ADAE

= ABAC

= 25

Solution.Since m(∠A) = m(∠A) and m(∠ADB) = m(∠AEC) then by the AA testwe have ∆ABD ∼ ∆ACE. Hence,

BD

CE=

AD

AE=

AB

AC=

6

15=

2

5

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Theorem 47.3 (SAS Test)If two sides of one triangle are proportional to the two corresponding sidesof the second triangle and the angles between the two sides of each triangleare equal then the two triangles are similar.

Proof.Let E ′ be a point on the line segement DE such that DE ′ = AB. Also, letF ′ be a point on DF such that DF ′ = AC. See Figure 47.3.

Figure 47.3

SinceAB

DE=

AC

DF

thenDE ′

DE=

DF ′

DF.

By Theorem 47.1, the lines EF and E ′F ′ are parallel. Thus, m(∠DEF ) =m(∠DE ′F ′) and m(∠DFE) = m(∠DF ′E ′). By the AA similarity theo-rem we have ∆DE ′F ′ ∼ ∆DEF. But by the SAS congruence test, we have∆ABC ∼= ∆DE ′F ′ and in particular ∆ABC ∼ ∆DE ′F ′ since two congruenttriangles are also similar. Hence, ∆ABC ∼ ∆DEF

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Theorem 47.4 (SSS Test)If the three sides of one triangle are proportional to the three correspondingsides of another triangle, then the two triangles are similar.

Proof.Let E ′ be a point on the line segement DE such that DE ′ = AB. Also, letF ′ be a point on DF such that DF ′ = AC. See Figure 47.4.

Figure 47.4

SinceAB

DE=

AC

DF=

BC

EF

thenDE ′

DE=

DF ′

DF=

BC

EF.

By Theorem 47.1, the lines EF and E ′F ′ are parallel. By SAS similaritytheorem , we have ∆DE ′F ′ ∼ ∆DEF. Thus,

DE ′

DE=

DF ′

DF=

E ′F ′

EF.

But DE′

DE= BC

EF. Hence, E ′F ′ = BC. By the SSS congruence theorem we have

∆ABC ∼= ∆DE ′F ′. Hence, ∆ABC ∼ ∆DE ′F ′. Since ∆DE ′F ′ ∼ ∆DEFthen ∆ABC ∼ ∆DEF

Example 47.3Show that the triangles in the figure below are similar.

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Solution.Since AB

DE= AC

DF= BC

EF= 2 then by Theorem 47.4 ∆ABC ∼ ∆DEF

Practice Problems

Problem 47.1Which of the following triangles are always similar?(a) right triangles(b) isosceles triangles(c) equilateral triangles

Problem 47.2Show that if ∆ABC ∼ ∆A′B′C ′ and ∆A′B′C ′ ∼ ∆A′′B′′C ′′ then ∆ABC ∼∆A′′B′′C ′′.

Problem 47.3Each pair of triangles is similar. By which test can they be proved to besimilar ?

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Problem 47.4Suppose ∆ABC ∼ ∆DEF with scaled factor k.(a) Compare the perimeters of the two triangles.(b) Compare the areas of the two triangles.

Problem 47.5Areas of two similar triangles are 144 sq.cm. and 81 sq.cm. If one side of thefirst triangle is 6 cm then find the corresponding side of the second triangle.

Problem 47.6The side of an equilateral triangle ∆ABC is 5 cm. Find the length of theside of another equilateral triangle ∆PQR whose area is four times area of∆ABC.

Problem 47.7The corresponding sides of two similar triangles are 4 cm and 6 cm. Findthe ratio of the areas of the triangles.

Problem 47.8A clever outdoorsman whose eye-level is 2 meters above the ground, wishes

261

to find the height of a tree. He places a mirror horizontally on the ground20 meters from the tree, and finds that if he stands at a point C which is4 meters from the mirror B, he can see the reflection of the top of the tree.How high is the tree?

Problem 47.9A child 1.2 meters tall is standing 11 meters away from a tall building. Aspotlight on the ground is located 20 meters away from the building andshines on the wall. How tall is the child’s shadow on the building?

Problem 47.10On a sunny day, Michelle and Nancy noticed that their shadows were dif-ferent lengths. Nancy measured Michelle’s shadow and found that it was 96inches long. Michelle then measured Nancy’s shadow and found that it was102 inches long.

(a) Who do you think is taller, Nancy or Michelle? Why?(b) If Michelle is 5 feet 4 inches tall, how tall is Nancy?(c) If Nancy is 5 feet 4 inches tall, how tall is Michelle?

Problem 47.11An engineering firm wants to build a bridge across the river shown below.An engineer measures the following distances: BC = 1,200 feet, CD = 40feet, and DE = 20 feet.

262

(a) Prove ∆ABC is similar to ∆EDC.(b) Railings cost $4 per foot. How much will it cost to put railings on bothsides of the bridge?

Problem 47.12At a certain time of the day, the shadow of a 5′ boy is 8′ long. The shadowof a tree at this same time is 28′ long. How tall is the tree?

Problem 47.13Two ladders are leaned against a wall such that they make the same anglewith the ground. The 10′ ladder reaches 8′ up the wall. How much furtherup the wall does the 18′ ladder reach?

Problem 47.14Given that lines DE and AB are parallel in the figure below, determine thevalue of x, i.e. the distance between points A and D.

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Problem 47.15In the figure below, lines AC and DE are vertical, and line CD is horizontal.Show that ∆ABC ∼ ∆EBD.

Problem 47.16Find a pair of similar triangles in each of these figures:

Problem 47.17Find x :

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Problem 47.18In the diagram, DE is parallel to AC. Also, BD = 4, DA = 6 and EC = 8.Find BC to the nearest tenth.

Problem 47.19Find BC.

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Problem 47.20Find BE

Problem 47.21Copy and complete the given table. It is given that OH

HJ= OI

IK.

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48 Solutions to Practice Problems

Problem 25.1Which of the following are integers?(a) −11 (b) 0 (c) 3

4(d) −9

3

Solution.(a), (b), and (d) are integers. Note that −9

3= −3. The number in (c) is the

decimal number 0.75

Problem 25.2Let N = {−1,−2,−3, · · ·}. Find(a) N ∪W(b) N ∩N, where N is the set of natural numbers or positive integers and Wis the set of whole numbers.

Solution.(a) N ∪W = Z.(b) N ∩ N = ∅

Problem 25.3What number is 5 units to the left of −95?

Solution.The number −100 is five units to the left of the number −95 on the numberline model

Problem 25.4A and B are 9 units apart on the number line. A is twice as far from 0 asB. What are A and B?

Solution.We use the method of guessing.

B A dist(A,O) dist(B,O) dist(A,O) = 2dist(B,O)?1 -8 8 1 NO2 -7 7 2 NO3 -6 6 3 YES

A second answer is B to be −3 and A to be 6. A third answer is A = 18 andB = 9

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Problem 25.5Represent −5 using signed counters and number line.

Solution.

Problem 25.6In terms of distance, explain why | − 4| = 4.

Solution.Because −4 is located 4 units to the left of zero

Problem 25.7Find the additive inverse (i.e. opposite) of each of the following integers.(a) 2 (b) 0 (c) −5 (d) m (e) −m

Solution.(a) −2 (b) 0 (c) 5 (d) −m (e) m

Problem 25.8Evaluate each of the following.(a) | − 5| (b) |10| (c) −| − 4| (d) −|5|

Solution.(a) | − 5| = 5 (b) |10| = 10 (c) −| − 4| = −4 (d) −|5| = −5

Problem 25.9Find all possible integers x such that |x| = 2.

Solution.We are looking for numbers that are 2 units away from zero. In this case,either x = −2 or x = 2

Problem 25.10Let W be the set of whole numbers, Z+ the set of positive integers (i.e.Z+ = N), Z− the set of negative integers, and Z the set of all integers. Findeach of the following.(a) W ∪ Z (b) W ∩ Z (c) Z+ ∪ Z− (d) Z+ ∩ Z− (e) W − Z+

268

Solution.(a) W ∪ Z = Z(b) W ∩ Z = W(c) Z+ ∪ Z− = {· · · ,−3,−2,−1, 1, 2, 3, · · ·}(d) Z+ ∩ Z− = ∅(e) W − Z+ = {0}

Problem 25.11What is the opposite or additive inverse of each of the following? (a and bare integers)(a) a + b (b) a− b

Solution.(a) −(a + b) = (−a) + (−b)(b) −(a− b) = (−a)− (−b) = (−a) + b

Problem 25.12What integer addition problem is shown on the number line?

Solution.(−4) + (−2) = −6

Problem 25.13(a) Explain how to compute −7 + 2 with a number line.(b) Explain how to compute −7 + 2 with signed counters.

Solution.

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Problem 25.14In today’s mail, you will receive a check for $86, a bill for $30, and anotherbill for $20. Write an integer addition equation that gives the overall gain orloss.

Solution.86 + (−30) + (−20)

Problem 25.15Compute the following without a calculator.(a) −54 + 25 (b) (−8) + (−17) (c) 400 + (−35)

Solution.(a) −54 + 25 = −(54− 25) = −29(b) (−8) + (−17) = −(8 + 17) = −25(c) 400 + (−35) = 400− 35 = 365

Problem 25.16Show two ways to represent the integer 3 using signed counters.

Solution.

Problem 25.17Illustrate each of the following addition using the signed counters.(a) 5 + (−3) (b) −2 + 3 (c) −3 + 2 (d) (−3) + (−2)

270

Solution.

Problem 25.18Demonstrate each of the additions in the previous problem using number linemodel.

271

Solution.

Problem 25.19Write an addition problem that corresponds to each of the following sentencesand then answer the question.(a) The temperature was −10◦C and then it rose by 8◦C. What is the newtemperature?(b) A certain stock dropped 17 points and the following day gained 10 points.What was the net change in the stock’s worth?(c) A visitor to a casino lost $200, won $100, and then lost $50. What is thechange in the gambler’s net worth?

Solution.(a) (−10) + 8 = −(10− 8) = −2◦C(b) (−17) + 10 = −(17− 10) = −7

272

(c) (−200)+100+(−50) = ((−200)+100)+(−50) = (−(200−100))+(−50) =(−100) + (−50) = −(100 + 50) = $− 150

Problem 25.20Compute each of the following: (a) |(−9) + (−5)| (b) |7 + (−5)| (c)|(−7) + 6|

Solution.(a) |(−9) + (−5)| = | − (9 + 5)| = | − 14| = 14(b) |7 + (−5)| = |7− 5| = |2| = 2(c) |(−7) + 6| = | − (7− 6)| = | − 1| = 1

Problem 25.21If a is an element of {−3,−2,−1, 0, 1, 2} and b is an element of {−5,−4,−3,−2,−1, 0, 1},find the smallest and largest values for the following expressions.(a) a + b (b) |a + b|

Solution.

a b a+b |a + b|-3 -5 -8 8-3 -4 -7 7-3 -3 -6 6-3 -2 -5 5-3 -1 -4 4-3 0 -3 3-3 1 -2 2

a b a+b |a + b|-2 -5 -7 7-2 -4 -6 6-2 -3 -5 5-2 -2 -4 4-2 -1 -3 3-2 0 -2 2-2 1 -1 1

a b a+b |a + b|-1 -5 -6 6-1 -4 -5 5-1 -3 -4 4-1 -2 -3 3-1 -1 -2 2-1 0 -1 1-1 1 0 0

a b a+b |a + b|0 -5 -5 50 -4 -4 40 -3 -3 30 -2 -2 20 -1 -1 10 0 0 00 1 1 1

a b a+b |a + b|1 -5 -4 41 -4 -3 31 -3 -2 21 -2 -1 11 -1 0 01 0 1 11 1 2 2

a b a+b |a + b|2 -5 -3 32 -4 -2 22 -3 -1 12 -2 0 02 -1 1 12 0 2 22 1 3 3

(a) The largest value of a + b is 3 and the smallest value is −8(b) The largest value of |a + b| is 8 and the smallest value is 0

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Problem 25.22Explain how to compute 5− (−2) using a number line.

Solution.

Problem 25.23Tell what subtraction problem each picture illustrates.

Solution.(a) (−5)− (−3) = −2(b) 2− 5 = −3(c) (−3)− 2 = −5

Problem 25.24Explain how to compute the following with signed counters.(a) (−6)− (−2) (b) 2− 6 (c) −2− 3 (d) 2− (−4)

274

Solution.

Problem 25.25(a) On a number line, subtracting 3 is the same as moving units tothe .(b) On a number line, adding −3 is the same as moving units to the

.

Solution.(a) On a number line, subtracting 3 is the same as moving three units to theleft.(b) On a number line, adding −3 is the same as moving three units to theleft

Problem 25.26An elevetor is at an altitude of −10 feet. The elevator goes down 30 ft.(a) Write an integer equation for this situation.(b) What is the new altitude?

Solution.(a) (−10)− 30(b) (−10)− 30 = (−10) + (−30) = −(10 + 30) = −40 ft

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Problem 25.27Compute each of the following using a− b = a + (−b).(a) 3− (−2) (b) −3− 2 (c) −3− (−2)

Solution.(a) 3− (−2) = 3 + [−(−2)] = 3 + 2 = 5(b) −3− 2 = (−3) + (−2) = −(3 + 2) = −5(c) −3− (−2) = (−3) + [−(−2)] = (−3) + 2 = −1

Problem 25.28Use number-line model to find the following.(a) −4− (−1) (b) −2− 1.

Solution.

Problem 25.29Compute each of the following.(a) |5− 11| (b) |(−4)− (−10)| (c) |8− (−3)| (d) |(−8)− 2|

Solution.(a) |5− 11| = | − 6| = 6(b) |(−4)− (−10)| = |6| = 6(c) |8− (−3)| = |11|11(d) |(−8)− 2| = | − 10| = 10

Problem 25.30Find x.(a) x + 21 = 16 (b) (−5) + x = 7 (c) x− 6 = −5 (d) x− (−8) = 17.

276

Solution.(a) x + 21 = 16 =⇒ x = 16− 21 = −5(b) (−5) + x = 7 =⇒ x = 7 + 5 = 12(c) x− 6 = −5 =⇒ x = 6− 5 = 1(d) x− (−8) = 17 =⇒ x = (−8) + 17 = 9

Problem 25.31Which of the following properties hold for integer subtraction. If a propertydoes not hold, disprove it by a counterexample.(a) Closure (b) Commutative (c) Associative (d) Identity

Solution.(a) Hold(b) Does not hold since 2− 3 6= 3− 2(c) Does not hold since (1− 2)− 3 6= 1− (2− 3)(d) Does not hold since 1− 0 6= 0− 1

Problem 26.1Use patterns to show that (−1)(−1) = 1.

Solution.First, we compute the following

(−1)× 3 = −3(−1)× 2 = −2(−1)× 1 = −1(−1)× 0 = 0

Thus, everytime the second factor decreases by 1 the results increase by 1.Hence, continuing the pattern we find

(−1)× 3 = −3(−1)× 2 = −2(−1)× 1 = −1(−1)× 0 = 0

(−1)× (−1) = 1

Problem 26.2Use signed counters to show that (−4)(−2) = 8.

277

Solution.

Problem 26.3Use number line to show that (−4)2 = −8.

Solution.

Problem 26.4Change 3× (−2) into a repeated addition and then compute the answer.

Solution.3× (−2) = (−2) + (−2) + (−2) = −6

Problem 26.5(a) Compute 4× (−3) with repeated addition.(b) Compute 4× (−3) using signed counters.(c) Compute 4× (−3) using number line.

Solution.(a) 4× (−3) = (−3) + (−3) + (−3) + (−3) = −12

278

Problem 26.6Show how (−2)× 4 can be found by extending a pattern in a whole-numbermultiplication.

Solution.We compute the following

3× 4 = 122× 4 = 81× 4 = 40× 4 = 0

It follows that each time the first factor is decreased by 1, the correspondingproduct is decreased by 4. Thus,

3× 4 = 122× 4 = 81× 4 = 40× 4 = 0

(−1)× 4 = −4(−2)× 4 = −8

Problem 26.7Mike lost 3 pounds each week for 4 weeks.(a) What was the total change in his weight?(b) Write an integer equation for this situation.

Solution.(a) He lost a total of 12 pounds.(b) (−3)× 4 = −12 pounds

Problem 26.8Compute the following without a calculator.(a) 3× (−8)(b) (−5)× (−8)× (−2)× (−3)

Solution.(a) 3× (−8) = −(3× 8) = −24(b) (−5)× (−8)× (−2)× (−3) = ((−5)× (−8))× ((−2)× (−3)) = (5× 8)×(2× 3) = 40× 6 = 240

279

Problem 26.9Extend the following pattern by writing the next three equations.

6× 3 = 186× 2 = 126× 1 = 66× 0 = 0

Solution.

6× 3 = 186× 2 = 126× 1 = 66× 0 = 0

6× (−1) = −66× (−2) = −126× (−3) = −18

Problem 26.10Find the following products.(a) 6(−5) (b) (−2)(−16) (c) −(−3)(−5) (d) −3(−7− 6).

Solution.(a) 6(−5) = −(6× 5) = −30(b) (−2)(−16) = 2× 16 = 32(c) −(−3)(−5) = −(3× 5) = −15(d) −3(−7− 6) = (−3)× (−13) = 3× 13 = 39

Problem 26.11Represent the following products using signed counters and give the results.(a) 3× (−2) (b) (−3)× (−4)

280

Solution.

Problem 26.12In each of the following steps state the property used in the equations.

a(b− c) = a[b + (−c)]= ab + a(−c)= ab + [−(ac)]= ab− ac

Solution.

a(b− c) = a[b + (−c)] adding the opposite= ab + a(−c) distributivity= ab + [−(ac)] Theorem 26.1(a)= ab− ac adding the opposite

Problem 26.13Extend the meaning of a whole number exponent


where a is any integer. Use this definition to find the following values.(a) (−2)4 (b) −24 (c) (−3)5 (d) −35

281

Solution.(a) (−2)4 = (−2)(−2)(−2)(−2) = 16(b) −24 = −(2× 2× 2× 2) = −16(c) (−3)5 = (−3)(−3)(−3)(−3) = 81(d) −35 = −(3× 3× 3× 3) = −81

Problem 26.14Illustrate the following products on an integer number line.(a) 2× (−5) (b) 3× (−4) (c) 5× (−2)

Solution.

Problem 26.15Expand each of the following products.(a) −6(x + 2) (b) −5(x− 11) (c) (x− 3)(x + 2)

Solution.(a) −6(x + 2) = −6x + (−6)(2) = −6x + (−12) = −6x− 12(b) −5(x− 11) = −5x− (−5)(11) = −5x− (−55) = −5x + 55(c) (x−3)(x+2) = x(x+2)−3(x+2) = x2+2x−(3x+6) = x2+2x−3x−6 =x2 − x− 6

Problem 26.16Name the property of multiplication of integers that is used to justify eachof the following equations.(a) (−3)(−4) = (−4)(−3)(b) (−5)[(−2)(−7)] = [(−5)(−2)](−7)

282

(c) (−5)(−7) is a unique integer(d) (−8)× 1 = −8(e) 4 · [(−8) + 7] = 4 · (−8) + 4 · 7

Solution.(a) Commutativty(b) Associativity(c) Closure(d) Identity element(e) Distributivity

Problem 26.17If 3x = 0 what can you conclude about the value of x?

Solution.Since 3x = 0 = 3(0) then by the Theorem 26.5 we have x = 0

Problem 26.18If a and b are negative and c is positive, determine whether the following arepositive or negative.(a) (−a)(−c) (b) (−a)(b) (c) (c− b)(c− a) (d) a(b− c)

Solution.(a) Since a is negative then −a is positive. Also, since c is positive then −cis negative. Thus, (−a)(−b) is negative.(b) Since −a is positive and b is negative then (−a)(b) is negative.(c) Since (c− b)(c− a) = c2 − ac− bc + ab = c2 + (−a)c + (−b)c + ab and c2

positive, (−a)c positive, (−b)c positive, and ab is positive then (c− b)(c− a)is positive.(d) a(b − c) = ab − ac = ab + (−a)c. But ab is positive as well as (−a)c.Hence, a(b− c) is positive

Problem 26.19Is the following equation true? a · (b · c) = (a · b) · (a · c).

Solution.False. Let a = 2, b = 3, and c = 4. Then a · (b · c) = 2 · (3 · 4) = 24 and(a · b) · (a · c) = (2 · 3) · (2 · 4) = 6 · 8 = 48

283

Problem 26.20Perform these divisions.(a) 36÷9 (b) (−36)÷9 (c) 36÷ (−9) (d) (−36)÷ (−9) (e) 165÷ (−11)(f) 275÷ 11

Solution.(a) If c = 36÷ 9 then 9c = 36. Consequently, c = 4.(b) If c = (−36)÷ 9 then 9c = −36 and consequently c = −4.(c) If c = 36÷ (−9) then −9c = 36 so that c = −4.(d) If c = (−36)÷ (−9) then −9c = −36 and so c = 4.(e) If c = 165÷ (−11) then −11c = 165 and so c = −15.(f) If c = 275÷ 11 then 11c = 275 and so c = 25

Problem 26.21Write another multiplication equation and two division equations that areequivalent to the equation

(−11) · (−25, 753) = 283, 283.

Solution.Multiplication equation: 11 · 25753 = 283283Division equation: 283283÷ (−11) = −25753Division equation: 283283÷ (−25753) = −11

Problem 26.22Write two multiplication equations and another division equation that areequivalent to the equation

(−1001)÷ 11 = −91.

Solution.Multiplication equation: 11× (−91) = −1001Multiplication equation: (−11)× 91 = −1001Division equation: 1001÷−11 = −91

Problem 26.23Use the definition of division to find each quotient, if possible. If a quotientis not defined, explain why.(a) (−40)÷ (−8) (b) (−143)÷ 11 (c) 0÷ (−5) (d) (−5)÷ 0 (e) 0÷ 0

284

Solution.(a) If c = (−40)÷ (−8) then −8c = −40 and so c = 5.(b) If c = (−143)÷ 11 then 11c = −143 and so c = −13.(c) If c = 0÷ (−5) then −5c = 0 and so c = 0.(d) If c = (−5) ÷ 0 then 0 × c = −5. There is no such c. This says that(−5)÷ 0 is undefined.(e) If c = 0 ÷ 0 then 0 × c = 0. Any integer works for c. By the uniquenessof c the division is undefined

Problem 26.24Find all integers x (if possible) that make each of the following true.(a) x ÷ 3 = −12 (b) x ÷ (−3) = −2 (c) x ÷ (−x) = −1 (d) 0 ÷ x = 0(e) x÷ 0 = 1

Solution.(a) If x÷ 3 = −12 then x = (−12)(3) = −36.(b) If x÷ (−3) = −2 then x = (−2)(−3) = 8.(c) If x÷ (−x) = −1 then x = (−1)(−x) = x so any integer works for x.(d) If 0÷ x = 0 then x× 0 = 0. Any integer works for x.(e) If x÷ 0 = 1 then x = 0

Problem 26.25Write two division equations that are equivalent to 3× (−2) = −6.

Solution.(i) (−6)÷ (−2) = 3 (ii) (−6)÷ 3 = −2

Problem 26.26Explain how to compute −10÷ 2 using signed counters.

Solution.Take a set of 10 negative counters and divide them into 2 equal groups offive negative counters each. So (−10)÷ 2 = −5

Problem 26.27Rewrite each of the following as an equivalent multiplication problem, andgive the solution.(a) (−54)÷ (−6) (b) 32÷ (−4)

285

Solution.(a) If c = (−54)÷ (−6) then −6c = −54 and so c = 9.(b) If c = 32÷ (−4) then −4c = 32 and so c = −8

Problem 26.28A store lost $480,000 last year.(a) What was the average net change per month?(b) Write an integer equation for this situation.

Solution.(a) $ - 40,000.(b) (−480, 000)÷ 12 = −40, 000

Problem 26.29Compute the following, using the correct rules for order of operations.(a) −22 − 3 (b) −5 + (−4)2 × (−2)

Solution.(a) −22 − 3 = −4− 3 = −7(b) −5 + (−4)2 × (−2) = −5 + 16× (−2) = −5− 32 = −37

Problem 26.30A stock change as follows for 5 days:-2,4,6,3,-1. What is the average dailychange in price?

Solution.We have (−2)+4+6+3+(−1) = 10 so the average daily change is 10÷5 = 2

Problem 26.31Compute: −2÷ (−2) + (−2)− (−2)

Solution.

−2÷ (−2) + (−2)− (−2) = 1 + (−2) + 2 = 1

Problem 26.32For what integers a and b does a÷ b = b÷ a?

Solution.Either a = b or a = −b

286

Problem 26.33Find each quotient, if possible.(a) [144÷ (−12)]÷ (−3) (b) 144÷ [−12÷ (−3)]

Solution.(a) [144÷ (−12)]÷ (−3) = (−12)÷ (−3) = 4(b) 144÷ [−12÷ (−3)] = 144÷ 4 = 36

Problem 26.34Compute the following writing the final answer in terms of positive exponents.(a) 4−2 · 46

(b) 63

6−7

(c) (3−4)−2

Solution.(a) 4−2 · 46 = 4−2+6 = 44

(b) 63

6−7 = 63−(−7) = 610

(c) (3−4)−2 = 3(−4)(−2) = 38

Problem 26.35Express each of the following in scientific notation.(a) 0.0004 (b) 0.0000016 (c) 0.000000000000071

Solution.(a) 0.0004 = 4× 10−4

(b) 0.0000016 = 1.6× 10−6

(c) 0.000000000000071 = 7.1× 10−14

Problem 26.36Hair on the human body can grow as fast as 0.0000000043 meter per second.(a) At this rate, how much would a strand of hair grow in one month of 30days? Express your answer in scientific notation.(b) About how long would it take for a strand of hair to grow to be 1 meterin length?

Solution.(a) 0.0000000043× 30× 24× 3600 = 0.0111456 = 1.11456× 10−2 meters.(b) It takes 1

0.0000000043≈ 232558139.534 seconds or 2.32558139534× 108

287

Problem 26.37Compute each of these to three significant figures using scientific notation.(a) (2.47× 10−5) · (8.15× 10−9)(b) (2.47× 10−5)÷ (8.15× 10−9)

Solution.(a) (2.47 × 10−5) · (8.15 × 10−9) = 2.47 × 8.15 × 10−14 ≈ 20.131 × 10−14 =2.0131× 10−13

(b) (2.47 × 10−5) ÷ (8.15 × 10−9) = (2.47 ÷ 8.15) × 104 ≈ 0.3030 × 104 =3.030× 105

Problem 26.38Convert each of the following to standard notation.(a) 6.84× 10−5 (b) 3.12× 107

Solution.(a) 6.84× 10−5 = 0.0000684(b) 3.12× 107 = 31200000

Problem 26.39Write each of the following in scientific notation.(a) 413,682,000 (b) 0.000000231 (c) 100,000,000

Solution.(a) 413, 682, 000 = 4.13682× 108

(b) 0.000000231 = 2.31× 10−7

(c) 100, 000, 000 = 1.0× 108

Problem 26.40Evaluate each of the following.(a) −52 + 3(−2)2

(b) −2 + 3 · 5− 1(c) 10− 3 · 7− 4(−2)÷ 2 + 3

Solution.(a) −52 + 3(−2)2 = −25 + 3× 4 = −25 + 12 = −13(b) −2 + 3 · 5− 1 = −2 + 15− 1 = 12(c) 10− 3 · 7− 4(−2)÷ 2 + 3 = 10− 21 + 4 + 3 = −4

288

Problem 26.41Evaluate each of the following, if possible.(a) (−10÷ (−2))(−2)(b) (−10 · 5)÷ 5(c) −8÷ (−8 + 8)(d) (−6 + 6)÷ (−2 + 2)(e) | − 24| ÷ 4 · (3− 15)

Solution.(a) (−10÷ (−2))(−2) = 5(−2) = −10(b) (−10 · 5)÷ 5 = −50÷ 5 = −10(c) −8÷ (−8 + 8) = −8÷ 0 is undefined(d) (−6 + 6)÷ (−2 + 2) = 0÷ 0 is undefined(e) | − 24| ÷ 4 · (3− 15) = 24÷ 4 · (−12) = 6 · (−12) = −72

Problem 26.42Use the number-line approach to verify each of the following.(a) −4 < 1 (b) −4 < −2 (c) −1 > −5

Solution.(a) −4 < 1 since −4 is to the left of 1.(b) −4 < −2 since − is to the left of −2.(c) −1 > −5 since −1 it to the right of −5

Problem 26.43Order each of the following lists from smallest to largest.(a) {−4, 4,−1, 1, 0}(b) {23,−36, 45,−72,−108}

Solution.(a) −4 < −1 < 0 < 1 < 4(b) −108 < −72 < −36 < 23 < 45

Problem 26.44Replace the blank by the appropriate symbol.(a) If x > 2 then x + 4 6(b) If x < −3 then x− 6 − 9

289

Solution.(a) If x > 2 then x + 4 > 6(b) If x < −3 then x− 6 < −9

Problem 26.45Determine whether each of the following statements is correct.(a) −3 < 5 (b) 6 < 0 (c) 3 ≤ 3 (d) −6 > −5 (e) 2× 4− 6 ≤ −3× 5 + 1

Solution.(a) True since −3 is to the left of 5 on the number line.(b) False since 6 is to the right of 0 on the number line.(c) True since 3 is equal to 3.(d) False since −6 is to the left of −5 on the number line.(e) False since the left-hand side is positive and the right-hand side is nega-tive

Problem 26.46What different looking inequality means the same as a < b?

Solution.

b > a

Problem 26.47Use symbols of inequalities to represent ”at most” and ”at least”.

Solution.”at most:” ≤”at least:” ≥

Problem 26.48For each inequality, determine which of the numbers −5, 0, 5 satisfies theinequality.(a) x > −5 (b) 5 < x (c) −5 > x

Solution.(a) 0 and 5(b) None(c) None

290

Problem 26.49Write the appropriate inequality symbol in the blank so that the two inequal-ities are true.(a) If x ≤ −3 then −2x 6(b) If x + 3 > 9 then x 6

Solution.(a) If x ≤ −3 then −2x ≥ 6(b) If x + 3 > 9 then x > 6

Problem 26.50How do you know when to reverse the direction of an inequality symbol?

Solution.Whenever the inequality is multiplied by a negative number

Problem 26.51Show that each of the following inequality is true by using the additionapproach.(a) −4 < −2 (b) −5 < 3 (c) −17 > −23

Solution.(a) −2 = (−4) + 2(b) 3 = (−5) + 8(c) −17 = (−23) + 6

Problem 26.52A student makes a connection between debts and negative numbers. So thenumber −2 represents a debt of $2. Since a debt of $10 is larger than a debtof $5 then the student writes −10 > −5. How would you convince him/herthat this inequality is false?

Solution.It is a true that a debt of $ 10 is larger than a debt of $ 5. However, ifwe use the number line model we see that −10 is to the left of −5 so that−10 < −5

Problem 26.53At an 8% sales tax rate, Susan paid more than $1500 sales tax when shepurchased her new Camaro. Describe this situation using an inequality withp denoting the price of the car.

291

Solution.

0.08p ≥ 1500

Problem 26.54Show that if a and b are positive integers with a < b then a2 < b2. Does thisresult hold for any integers?

Solution.Multiplying a < b by a and b respectively and using the fact that bothnumbers are positive we obtain a2 < ab and ab < b2. By transitivity, we havea2 < b2. This is false if both a and b are not positive. For example,−2 < 1but (−2)2 > 12

Problem 26.55Elka is planning a rectangular garden that is twice as long as it is wide. Ifshe can afford to buy at most 180 feet of fencing, then what are the possiblevalues for the width?

Solution.Let L denote the length and W denote the width. Then L = 2W. Accordingto the problem, we have 2L+2W ≤ 180. Thus, 4W +2W ≤ 180 or 6W ≤ 180.Hence, W ≤ 30

Problem 27.1Show that each of the following numbers is a rational number.(a) −3 (b) 41

2(c) −5.6 (d) 25%

Solution.(a) −3 = −3

1

(b) 412

= 4 + 12

= 92

(c) −5.6 = −5610

= −285

(d) 25% = 25100

= 14

Problem 27.2Which of the following are equal to −3?

−3

1,

3

−1,3

1,−3

1,−3

−1,−−3

1,−−3

−1.

292

Solution.

−3 =−3

1=

3

−1= −3

1= −−3

−1

Problem 27.3Determine which of the following pairs of rational numbers are equal.(a) −3

5and 63

−105

(b) −18−24

and 4560

.

Solution.(a) Since (−3)× (−105) = 5× 63 then −3

5= 63

−105

(b) Since (−24)× 45 = (−18)× 60 then −18−24

= 4560

Problem 27.4Rewrite each of the following rational numbers in simplest form.(a) 5

−7(b) 21

−35(c) −8

−20(d) −144

180

Solution.(a) Already in simpest form.(b) 21

−35= 3×7

(−5)×7= 3

−5

(c) −8−20

= 2×(−4)5×(−4)

= 25

(d) −144180

= −24×32

22×32×5= −22

5= −4

5

Problem 27.5How many different rational numbers are given in the following list?

2

5, 3,

−4

−10,39

13,7

4

Solution.

2

56= 3 6= 7

4

Problem 27.6Find the value of x to make the statement a true one.

(a) −725

= x500

(b) 183

= −5x

293

Solution.(a) −7

25= −140

500so x = −140

(b) 6 = −5x

so x = −56

Problem 27.7Find the prime factorizations of the numerator and the denominator and usethem to express the fraction 247

−77in simplest form.

Solution.Fraction is already in simplest form since

247

−77=

19× 13

(−7)× 11

Problem 27.8(a) If a

b= a

c, what must be true?

(b) If ac

= bc, what must be true?

Solution.(a) b = c(b) a = b

Problem 27.9Use number line model to illustrate each of the following sums.(a) 3

4+ −2

4(b) −3

4+ 2

4(c) −3

4+ −1

4

Solution.

294

Problem 27.10Perform the following additions. Express your answer in simplest form.(a) 6

8− −25

100(b) −57

100+ 13

10

Solution.(a) 6

8− −25

100= 3

4+ 1

4= 4

4= 1

(b) −57100

+ 1310

= −57100

+ 130100

= 73100

Problem 27.11Perform the following subtractions. Express your answer in simplest form.(a) 137

214− −1

3(b) −23

100− 198

1000

Solution.(a) 137

214− −1

3= 411

642+ 214

624= 625

624

(b) −23100

− 1981000

= −2301000

− 1981000

= −4281000

Problem 27.12Compute the following differences.(a) 2

3− −9

8(b)

(−21

4

)− 42

3

Solution.(a) 2

3− −9

8= 16

24+ 27

24= 43

24

(b)(−21

4

)− 42

3= −9

4− 14

3= −27

12− 56

12= −83

12

Problem 27.13Multiply the following rational numbers. Write your answers in simplestform.

(a) 35· −10

21(b) −6

11· −33

18(c) 5

12· 48−15

· −98

Solution.(a) 3

5· −10

21= 3

5· (−2)·5

3·7 = −27

(b) −611· −33

18= −6

11· (−3)·11

6·3 = 1

(c) 512· 48−15

· −98

= 512· 4·12

(−3)·5 ·(−3)·3

4·2 = 32

Problem 27.14Find the following quotients. Write your answers in simplest form.

(a) −89÷ 2

9(b) 12

15÷ −4

3(c) −13

24÷ −39

−48

295

Solution.(a) −8

9÷ 2

9= −8

9· 9

2= −4

(b) 1215÷ −4

3= 4

3· 3−4

= −1

(c) −1324÷ −39

−48= −13

24· 16

13= −24

16= −3

2

Problem 27.15State the property that justifies each statement.

(a)(

57· 7

8

)· −8

3= 5

7·(

78· −8

3

)(b) 1

4

(83

+ −54

)= 1

4· 8

3+ 1

4· −5

4

Solution.(a) Associativity(b) Distributivity

Problem 27.16Compute the following and write your answers in simplest form.

(a) −4027÷ −10

9(b) 21

25÷ −3

5(c) −10

9÷ −9

8

Solution.(a) −40

27÷ −10

9= −40

27· 9−10

= 43

(b) 2125÷ −3

5= 21

25· 5−3

= −75

(c) −109÷ −9

8= −10

9· 8−9

= 8081

Problem 27.17Find the reciprocals of the following rational numbers.

(a) 4−9

(b) 0 (c) −32

(d) −4−9

Solution.(a) −9

4

(b) Does not exist(c) 2

−3

(d) −9−4

= 94

Problem 27.18Compute:

(−47· 2−5

)÷ 2

−7.

296

Solution. (−47· 2−5

)÷ 2

−7= 8

35÷ 2

−7

= 835· −7

2

= −45

Problem 27.19If a

b· −4

7= 2

3what is a

b?

Solution.

a

b=

2

3÷ −4

7=

2

3· 7

−4= −7

6


2×−52

3

Solution.

−41

2×−5

2

3=−9

2· −17

3=

51

2


9÷ 510

11

Solution.

−178

9÷ 5

10

11=−161

9÷ 65

11=−161

9· 11

65= −1771

585

Problem 27.22Compute each of the following:

(a) −(

34

)2(b)

(−3

4

)2(c)

(34

)2 ·(

34

)7

Solution.(a) −

(34

)2= − 9

16

(b)(−3

4

)2= 9

16

(c)(

34

)2 ·(

34

)7=

(34

)2+7=

(34

)9

Problem 27.23True or false: (a) −2

3< −3

7(b) 15

−9> −13

4.

297

Solution.(a) First we reduce to same denominator: −2

3= −14

21and −3

7= −9

21. Since

−1421

< −921

then the given statement is true.(b) We have 15

−9= −60

36and −13

4= −96

36. Since −96

36< −60

36then the statement is

true


4< −1

4using the addition approach.

Solution.Since −3

4+ 2

4= −1

4then −3

4< −1

4


4< −1

4by using a number line.

Solution.

Problem 27.26Put the appropriate symbol, <, =, > between each pair of numbers to makea true statement.

(a) −56

− 1112

(b) −13

54

(c) −1215

36−45

(d) − 312

−420

Solution.(a) −5

6= −10

12> −11

12

(b) −13

< 54

(c) −1215

= −3645

= 36−45

(d) − 312

= −14

= − 520

< −420

298

Problem 27.27Find three rational numbers between 1

4and 2

5.

Solution.Since 1

4< 2

5then by the density property we have

1

4<

1

2

(1

4+

2

5

)<

2

5

or1

4<

1

2

(5

20+

8

20

)<

2

5

1

4<

13

40<

2

5

Similarly,1

4<

1

2

(1

4+

13

40

)<

13

40<

1

2

(13

40+

2

5

)<

2

5

Problem 27.28The properties of rational numbers are used to solve inequalities. For exam-ple,

x + 35

< −710

x + 35

+(−3

5

)< −7

10+

(−3

5

)x < −13

10

Solve the inequality

−2

5x +

1

5> −1.

Solution.Subtract 1

5from both sides

−2

5x > −1− 1

5= −6

5

Now multiply both sides by −52

to obtain

x < 3

299

Problem 27.29Solve each of the following inequalities.

(a) x− 65

< −127

(b) 25x < −7

8

(c) −37

x > 85

Solution.(a)

x− 65

< −127

x− 65

+ 65

< −127

+ 65

x < −1835

(b)25x < −7

852· 2

5x < 5

2· −7

8

x < −3516

(c)−37

x > 85

7−3· −3

7x < 7

−3· 8

5

x < −5615

Problem 27.30Verify the following inequalities.

(a) −45

< −34

(b) 110

< 14

(c) 19−60

> −13

Solution.(a) Since −4

5= −16

20and −3

4= −15

20then −4

5< −3

4

(b) Since 110

= 220

and 14

= 520

then 110

< 14

(c) Since −13

= −2060

then 19−60

> −13

Problem 27.31Use the number-line approach to arrange the following rational numbers inincreasing order:

(a) 45,−1

5, 2

5

(b) −712

, −23

, 3−4

300

Solution.

Problem 27.32Find a rational number between 5

12and 3

8.

Solution.

3

8<

1

2

(5

12+

3

8

)<

5

12

or3

8<

1

2

(10

24+

9

24

)<

5

12

3

8<

19

48<

5

12

Problem 27.33Complete the following, and name the property that is used as a justification.

(a) If −23

< 34

and 34

< 75

then −23

75.

(b) If −35

< −611

then(−3

5

)·(

23

) (−611

)·(

23

)(c) If −4

7< 7

4then −4

7+ 5

8< 7

4+

(d) If −34

> 113

then(−3

4

)·(−5

7

) (113

)·(−5

7

)(e) There is a rational number any two unequal rational numbers.

301

Solution.(a) If −2

3< 3

4and 3

4< 7

5then −2

3< 7

5. (Transitivity)

(b) If −35

< −611

then(−3

5

)·(

23

)<

(−611

)·(

23

). (Multiplication property)

(c) If −47

< 74

then −47

+ 58

< 74

+ 58

(Addition property)

(d) If −34

< 113

then(−3

4

)·(−5

7

)>

(113

)·(−5

7

)(Multiplication by a nega-

tive rational)

(e) There is a rational number between any two unequal rational numbers.(Density property)

Problem 28.1Given a decimal number, how can you tell whether the number is rational orirrational?

Solution.If the number is terminating or nonterminating and repeating then the dec-imal number is rational. Otherwise, the number is irrational

Problem 28.2Write the following repeating decimal numbers as a fraction.(a) 0.8 (b) 0.37 (c) 0.02714

Solution.(a) Let x = 0.8. Then 10x = 8.8 = 8 + x. Solving for x we find x = 8

9.

(b) Let x = 0.37. Then 100x = 37.37 = 37 + x. Solving for x we find x = 3799

(c) Let x = 0.02714. Then 100000x = 27140.02714 = 2714 + x. Solving for xwe find x = 2714

99999

Problem 28.3Which of the following describe 2.6?(a) A whole number(b) An integer(c) A rational number(d) An irrational number(e) A real number

302

Solution.(c) and (e). Note that 2.6 = 26

10= 13

5

Problem 28.4Classify the following numbers as rational or irrational.(a)

√11 (b) 3

7(c) π (d)

√16

Solution.(a) Irrational, nonterminating and nonrepeating(b) Rational(c) Irrational, nonterminating and nonrepeating(d)

√16 = 4, a rational

Problem 28.5Classify the following numbers as rational or irrational.(a) 0.34938661 · · · (b) 0.26 (c) 0.565665666 · · ·

Solution.(a) Irrational, nontermintaing and nonrepeating.(b) Rational, nonterminating and repeating.(c) Irrational, nonterminating and nonrepeating

Problem 28.6Match each word in column A with a word in column B.

A BTerminating RationalRepeating IrrationalInfinite nonrepeating

Solution.(Terminating, Rational), (Repeating, Rational), (Infinite nonrepeating, Irra-tional)

Problem 28.7(a) How many whole numbers are there between −3 and 3 (not including 3and −3)?(b) How many integers are there between −3 and 3?(c) How many real numbers are there between −3 and 3?

303

Solution.(a) 0, 1, 2(b) −2,−1, 0, 1, 2(c) infinite numbers

Problem 28.8Prove that

√3 is irrational.

Solution.Suppose that

√3 is rational. Then

√3 = a

bfor some integers a and b 6= 0. But

then a2

b2= 3 so that a2 = 4b2. If a has an even number of prime factors then

a2 has an even number of prime factors. If a has an odd number of primefactors then a2 has an even number of prime factors. So, a2 and b2 haveboth even number of prime factors. But 3 · b2 has an odd number of primefactors. So we have that a2 has an even number and an odd number of primefactors. This cannot happen by the Fundamental Theorem of Arithmeticwhich says that every positive integer has a unique number of prime factors.In conclusion,

√3 cannot be written in the form a

bso it is not rational

Problem 28.9Show that 1 +

√3 is irrational.

Solution.Suppose that 1+

√3 = c is rational. Then

√3 = c−1. Since both 1 and c are

rationals then c− 1 is also rational. This shows that√

3 is rational. But wejust proved in the previous problem that

√3 is irrational. Hence, c = 1+

√3

must be irrational

Problem 28.10Find an irrational number between 0.37 and 0.38

Solution.√2

3− 1

10

Problem 28.11Write an irrational number whose digits are twos and threes.

Solution.One answer is 0.232233222333 · · ·

304

Problem 28.12Classify each of the following statement as true or false. If false, give acounter example.

(a) The sum of any rational number and any irrational number is a rationalnumber.(b) The sum of any two irrational numbers is an irrational number.(c) The product of any two irrational numbers is an irrational number.(d) The difference of any two irrational numbers is an irrational number.

Solution.(a) False: 1 +

√2 is irrational.

(b) False:√

2 + (−√

2) = 0 which is rational.(c) False:

√2 ·√

2 = 2 which is rational.(d) False:

√2−

√2 = 0 which is rational

Problem 28.13Construct the lengths

√2,√

3,√

4,√

5, · · · as follows.(a) First construct a right triangle with both legs of length 1. What is thelength of the hypotenuse?(b) This hypotenuse is the leg of the next right triangle. The other leg haslength 1. What is the length of the hypotenuse of this triangle?(c) Continue drawing right triangles, using the hypotenuse of the proceedingtriangle as a leg of the next triangle until you have constructed one withlength

√6.

305

Solution.

Problem 28.14Arrange the following real numbers in increasing order.0.56, 0.56, 0.566, 0.56565556 · · ·, 0.566

Solution.

0.56 < 0.56565556 · · · < 0.56 < 0.566 < 0.566

Problem 28.15Which property of real numbers justify the following statement

2√

3 + 5√

3 = (2 + 5)√

3 = 7√

3

Solution.Distributivity of addition with respect to multiplication

306

Problem 28.16Find x : x + 2

√2 = 5

√2.

Solution.Subtract 2

√2 from both sides to obtain

x = 3√

2

Problem 28.17Solve the following equation: 2x− 3

√6 = 3x +

√6

Solution.

2x− 3√

6 = 3x +√

6

−2x−√

6 + 2x− 3√

6 = 3x +√

6− 2x−√

6

−4√

6 = x

Problem 28.18Solve the inequality: 3

2x− 2 < 5

6x + 1

3

Solution.

32x− 2 < 5

6x + 1

332x− 2− 5

6x + 2 < 5

6x + 1

3− 5

6x + 2

23x < 7

332· 2

3x < 3

2· 7

3

x < 72

Problem 28.19Determines for what real number x, if any, each of the following is true:(a)

√x = 8 (b)

√x = −8 (c)

√−x = 8 (d)

√−x = −8

Solution.(a)

√x = 8 =⇒ x = 82 = 64.

(b) Since√

x ≥ 0 then the given equation has no real solutions.(c)

√−x = 8 =⇒ −x = 64 =⇒ x = −64.

(d) Since√−x ≥ 0 then the given equation has no real solutions

Problem 28.20Express the following values without exponents.(a)25

12 (b) 32

15 (c) 9

52 (d) (−27)

43

307

Solution.(a)25

12 = 5

(b) 3215 = (25)

15 = 2

(c) 952 = (32)

52 = 35 = 405

(d) (−27)43 = [(−3)3]

43 = (−3)4 = 81

Problem 28.21Write the following radicals in simplest form if they are real numbers.(a) 3

√−27 (b) 4

√−16 (c) 5

√32

Solution.(a) 3

√−27 = −3

(b) 4√−16 undefined since the radicand is negative and the index is even.

(c) 5√

32 = 2

Problem 28.22A student uses the formula

√a√

b =√

ab to show that -1 = 1 as follows:

−1 = (√−1)2 =

√−1√−1 =

√(−1)(−1) =

√1 = 1.

What’s wrong with this argument?

Solution.√−1 is not a real number since

√a is a real number when a ≥ 0

Problem 28.23Give an example where the following statement is true:

√a + b =

√a +

√b

Solution.a = b = 0 is such an example. Another example is a = 1, b = 0

Problem 28.24Give an example where the following statement is false:

√a + b =

√a +

√b.

Solution.Take a = b = 1. In this case.

√1 + 1 =

√2 6=

√1 +

√1 = 2

Problem 28.25Find an example where the following statement is false:

√a2 + b2 = a + b.

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Solution.Take a = b = 1 and see the previous solution

Problem 28.26Express the following values without using exponents:(a) (310)

35 (b) 81−

54

Solution.(a) (310)

35 = 310· 3

5 = 36 = 729

(b) 81−54 = (34)−

54 = 34·−5

4 = 3−5 = 181

Problem 28.27If possible, find the square root of each of the following without using acalculator.(a) 225 (b) 169 (c) -81 (d) 625

Solution.(a) 225 = 152 so

√225 = 15

(b) 169 = 132 so√

169 = 13(c)

√−81 is undefined since the radicand is negative.

(d) 625 = 252 so√

625 = 25

Problem 28.28Write each of the following in the form a

√b where a and b are integers and

b has the least value possible.(a)

√180 (b)

√363 (c)

√252

Solution.(a)

√180 =

√36 · 5 = 6

√5

(b)√

363 =√

3 · 112 = 11√

3(c)

√252 =

√22 · 32 · 7 = 6

√7

Problem 29.1Plot the points whose coordinates are given on a Cartesian coordinate system.

(a) (2, 4), (0,−3), (−2, 1), (−5,−3).(b) (−3,−5), (−4, 3), (0, 2), (−2, 0).

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Solution.

Problem 29.2Plot the following points using graph papers.(a) (3,2) (b) (5,0) (c) (0,-3) (d) (-3,4) (e) (-2,-3) (f) (2,-3)

Solution.


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(x,y) x > 0, y > 0 x < 0, y > 0 x > 0, y < 0 x < 0, y < 0Quadrant

Solution.

(x,y) x > 0, y > 0 x < 0, y > 0 x > 0, y < 0 x < 0, y < 0Quadrant I II IV III

Problem 29.4In the Cartesian plane, shade the region consisting of all points (x, y) thatsatisfy the two conditions

−3 ≤ x ≤ 2 and 2 ≤ y ≤ 4

Solution.

Problem 29.5Determine which of the following graphs represent a functions.

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Solution.(a) and (d) represent functions. (b) and (c) fail the vertical line test

Problem 29.6Consider the function f whose graph is given below.

(a) Complete the following table

x -2 -1 0 1 2 3f(x)

(b) Find the domain and range of f.(c) For which values of x is f(x) = 2.5?

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Solution.(a)

x -2 -1 0 1 2 3f(x) 0 3 4 1 3 5

(b) Dom(f) = R, and Range(f) = R.(c) We have to solve the two equations −x2 + 4 = 2.5 and 2x − 1 = 2.5.Solving the first equation we find x2 = 4−2.5 = 1.5 so x =

√1.5 and solving

the second equation we find x = 3.52

= 1.75

Problem 29.7Make a table of five values of the function f(x) = 2x + 3 and then use thepoints to sketch the graph of f(x).

Solution.

x -2 -1 0 1 2f(x) -1 1 3 5 7

Problem 29.8Make a table of five values of the function f(x) = 1

2x2 + x and then use the

points to sketch the graph of f(x).

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Solution.

x -2 -1 0 1 2f(x) 0 -1/2 0 3/2 4

Problem 29.9Make a table of five values of the function f(x) = 3x and then use the pointsto sketch the graph of f(x).

Solution.

x -2 -1 0 1 2f(x) 1/9 1/3 1 3 9

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Problem 29.10Make a table of five values of the function f(x) = 2 − x3 and then use thepoints to sketch the graph of f(x).

Solution.

x -2 -1 0 1 2f(x) 10 3 2 1 -6

Problem 29.11Which type of function best fits each of the following graphs: linear, quadratic,cubic, exponential, or step?

Solution.(a) Exponential (b) Cubic (c) Quadratic

Problem 29.12Suppose that a function f is given by a table. If the output changes by afixed amount each time the input changes by a constant then the function islinear. Determine whether each of the following functions below are linear.

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x 0 2 4 6f(x) 20 40 80 160

x 10 20 30 40g(x) 6 12 18 24

Solution.(a) f is not linear since every time the independent variable increases by 2the dependent variable is not changing by a fixed amount.(b) g is linear since every time x increases by 10, g increases by 6

Problem 29.13Show how to solve the equation 2x + 3 = 11 using a calculator.

Solution.You graph the two lines Y1 = 2x + 3 and Y2 = 11 on the same coordinatesystem and then you find the point of intersection which is the answer to theproblem

Problem 29.14In the linear function f(x) = mx + b the parameter m is called the slope.The slope of the line determines whether the line rises, falls, is vertical orhorizontal. Classify the slope of each line as positive, negative, zero, orundefined.

Solution.(a) Negative (b) Zero (c) Positive (d) Undefined

Problem 29.15Algebraically, one finds the slope of a line given two points (x1, y1) and (x2, y2)on the line by using the formula

y2 − y1

x2 − x1

.

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Would the ratioy1 − y2

x1 − x2

give the same answer? Explain.

Solution.Yes since one obtains an equivalent fraction by multiplying the numeratorand the denominator by −1

Problem 29.16Water is being pumped into a tank. Reading are taken every minutes.

Time(min) 0 3 6 9 12Quarts of water 0 90 180 270 360

(a) Plot the 5 points.(b) What is the slope of the line joining the 5 points?(c) Estimate how much water is in the tank after 1 minute.(d) At what rate is the water being pumped in?

Solution.

(a)(b) 90−0

3−0= 30

(c) Since the equation of the line is Q(t) = 30t then Q(1) = 30 quarts.(d) 30 quarts per minute

Problem 29.17(a) Graph f(x) = x2 − 2 by plotting the points x = −2,−1, 0, 1, 2.(b) How is this graph related to the graph of y = x2?


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x -2 -1 0 1 2f(x) 2 -1 -2 -1 2

(b) A vertical translation of the graph of y = x2 downward by two units

Problem 29.18(a) Graph f(x) = x2 + 3 by plotting the points x = −2,−1, 0, 1, 2.(b) How is this graph related to the graph of y = x2?


x -2 -1 0 1 2f(x) 7 4 3 4 7

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(b) A vertical translation of the graph of y = x2 upward by three units

Problem 29.19(a) Graph f(x) = (x− 1)2 by plotting the points x = −2,−1, 0, 1, 2.(b) How is this graph related to the graph of y = x2?


x -2 -1 0 1 2f(x) 9 4 1 0 1

(b) A horizontal translation of the graph of y = x2 to the right by one unit

Problem 29.20(a) Graph f(x) = (x + 2)2 by plotting the points x = −2,−1, 0, 1, 2.(b) How is this graph related to the graph of y = x2?


x -3 -2 -1 0 1f(x) 1 0 1 4 9

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(b) A horizontal translation of the graph of y = x2 to the left by two units

Problem 29.21Graph each equation by plotting points that satisfy the equation.

(a) x− y = 4.(b) y = −2|x− 3|.(c) y = 1

2(x− 1)2.


x -2 -1 0 1 2f(x) -6 -5 -4 -3 -2

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(b) We have

x 0 1 2 3 4f(x) -6 -4 -2 0 -2

(c) We have

x -2 -1 0 1 2f(x) 4.5 2 0.5 0 0.5

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Problem 29.22Find the x- and y-intercepts of each equation.

(a) 2x + 5y = 12.(b) x = |y| − 4.(c) |x|+ |y| = 4.

Solution.(a) x-intercept: y = 0 =⇒ 2x = 12 =⇒ x = 6.y-intercept: x = 0 =⇒ 5y = 12 =⇒ y = 12

5.

(b) x-intercept: y = 0 =⇒ x = −4.y-intercept: x = 0 =⇒ |y| = 4 =⇒ y = ±4.(c) x-intercept: y = 0 =⇒ |x| = 4 =⇒ x = ±4.y-intercept: x = 0 =⇒ |y| = 4 =⇒ y = ±4

Problem 30.1Following are the ages of the 30 students at Washington High School whoparticipated in the city track meet. Draw a line plot to represent these data.

10 10 11 10 13 8 10 13 14 914 13 10 14 11 9 13 10 11 1211 12 14 13 12 8 13 14 9 14

Solution.

Problem 30.2The height (in inches) of the players on a professional basketball team are70, 72, 75, 77, 78, 78, 80, 81,81,82, and 83. Make a line plot of the heights.

Solution.

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Problem 30.3Draw a line Plot for the following dataset.

50 35 70 55 50 30 4065 50 75 60 45 35 7560 55 55 50 40 55 50

Solution.

Problem 30.4Given below the score of a 26 fourth graders.

64 82 85 99 96 81 97 80 81 80 84 87 9875 86 88 82 78 81 86 80 50 84 88 83 82

Make a stem-and-leaf display of the scores.

Solution.

5 06 47 5 88 0 0 0 1 1 1 2 2 2 3 4 4 5 6 6 7 8 89 6 7 8 9

Problem 30.5Each morning, a teacher quizzed his class with 20 geography questions. Theclass marked them together and everyone kept a record of their personalscores. As the year passed, each student tried to improve his or her quizmarks. Every day, Elliot recorded his quiz marks on a stem and leaf plot.This is what his marks looked like plotted out:

0 3 6 51 0 1 4 3 5 6 5 6 8 9 7 92 0 0 0 0

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What is his most common score on the geography quizzes? What is hishighest score? His lowest score? Are most of Elliot’s scores in the 10s, 20sor under 10?

Solution.20 is his most common score. His highest score is 20. His lowest score is 3.His most scores are in the 10s

Problem 30.6A teacher asked 10 of her students how many books they had read in thelast 12 months. Their answers were as follows:

12, 23, 19, 6, 10, 7, 15, 25, 21, 12

Prepare a stem and leaf plot for these data.

Solution.

0 6 71 0 2 2 5 92 1 3 5

Problem 30.7Make a back-to-back stem and leaf plot for the following test scores:

Class 1:100 96 93 92 92 92 90 9089 89 85 82 79 75 74 7373 73 70 69 68 68 65 6135

Class 2:79 85 56 79 84 64 44 5769 85 65 81 73 51 61 6771 89 69 77 82 75 89 9274 70 75 88 46

Solution.

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5 34 4 65 1 6 7

9 8 8 5 1 6 1 4 5 7 9 99 5 4 3 3 3 0 7 0 1 3 4 5 5 7 9 9

9 9 5 2 8 1 2 4 5 5 8 9 96 3 2 2 2 0 0 9 2

0 10

Problem 30.8Suppose a sample of 38 female university students was asked their weightsin pounds. This was actually done, with the following results:

130 108 135 120 97 110130 112 123 117 170 124120 133 87 130 160 128110 135 115 127 102 13089 135 87 135 115 110105 130 115 100 125 120120 120

(a) Suppose we want 9 class intervals. Find CW.(b) Construct a frequency distribution.(c) Construct the corresponding histogram.

Solution.(a) CW = 170−87

9≈ 10

(b)

Class Frequency80 - 90 491 - 101 2102 - 112 7113 - 123 10124 - 134 10135 - 145 4146 - 156 0157 - 167 1168 - 178 1

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(c)

Problem 30.9The table below shows the response times of calls for police service measuredin minutes.

34 10 4 3 9 18 43 14 8 15 19 24 936 5 7 13 17 22 273 6 11 16 21 26 3132 38 40 30 47 53 146 12 18 23 28 333 4 62 24 35 5415 6 13 19 3 44 20 5 4 5 510 25 7 7 42 44


Solution.We will use 6 class intervals so that CW = 62−3

6= 9.8 ≈ 10. Thus, a possible

frequency distribution is the following.

Class Frequency1 - 11 2712 - 22 1623 - 33 1034 - 44 845 - 55 356 - 66 1

The corresponding histogram is given by the figure below.

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Problem 30.10A nutritionist is interested in knowing the percent of calories from fat whichAmericans intake on a daily basis. To study this, the nutritionist randomlyselects 25 Americans and evalustes the percent of calories from fat consumedin a typical day. The results of the study are as follows

34% 18% 33% 25% 30%42% 40% 33% 39% 40%45% 35% 45% 25% 27%23% 32% 33% 47% 23%27% 32% 30% 28% 36%


Solution.Constructing 3 class intervals each of width CW = 47−18

3= 9.8 ≈ 10. The

frequency distribution chart is

Class Frequency8 - 28 829 - 39 1140 - 50 6

The corresponding histogram is given by the figure below.

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Problem 30.11Given are several gasoline vehicles and their fuel consumption averages.

Buick 27 mpgBMW 28 mpgHonda Civic 35 mpgGeo 46 mpgNeon 38 mpgLand Rover 16 mpg

(a) Draw a bar graph to represent these data.(b) Which model gets the least miles per gallon? the most?

Solution.(a)The bar graph is given by the figure below.

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(b) Land Rover gets the least mileage whereas Geo gets the most

Problem 30.12The bar chart below shows the weight in kilograms of some fruit sold oneday by a local market.

How many kg of apples were sold? How many kg of oranges were sold?

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Solution.52 kilograms of apples were sold and 40 kg of oranges were sold

Problem 30.13The figures for total population at decade intervals since 1959 are givenbelow:

Y ear Total UK Resident Population1959 51, 956, 0001969 55, 461, 0001979 56, 240, 0001989 57, 365, 0001999 59, 501, 000

Construct a bar chart for this data.

Solution.

Problem 30.14The following data gives the number of murder victims in the U.S in 1978 clas-sified by the type of weapon used on them. Gun, 11, 910; cutting/stabbing,3, 526; blunt object, 896; strangulation/beating, 1, 422; arson, 255; all others705. Construct a bar chart for this data. Use vertical bars.

330

Solution.

Problem 30.15The table below shows the ingredients used to make a sausage and mushroompizza.

Ingredient %Sausage 7.5Cheese 25Crust 50

TomatoSause 12.5Mushroom 5

Plot a pie chart for this data.

Solution.First we find the measure of the central angle of each:

Sausage 7.5× 3.6 = 27Cheese 25× 3.6 = 90Crust 50× 3.6 = 180Tomato Sauce 12.5× 3.6 = 45Mushroom 5× 3.6 = 18

The pie chart is given below.

331

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Problem 30.16A newly qualified teacher was given the following information about theethnic origins of the pupils in a class.

Ethnic origin No. of pupilsWhite 12Indian 7

BlackAfrican 2Pakistani 3

Bangladeshi 6TOTAL 30

Plot a pie chart representing the data.

Solution.First we find the measure of the central angle of each:

White (12÷ 30)× 360 = 144Indian (7÷ 30)× 360 = 82.8Black African (2÷ 30)× 360 = 25.2Pakistani (3÷ 30)× 360 = 36Bangadeshi (6÷ 30)× 360 = 72


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Problem 30.17The following table represent a survey of people’s favorite ice cream flavor

Flovor Number of peopleV anilla 21.0%

Chocolate 33.0%Strawberry 12.0%Raspberry 4.0%

Peach 7.0%Neopolitan 17.0%

Other 6.0%

Plot a pie chart to represent this data.

Solution.First we find the measure of the central angle of each category:

Vanilla 21.0× 3.6 = 75.6Chocolate 33.0× 3.6 = 118.8Strawberry 12.0× 3.6 = 43.2Raspberry 4.0× 3.6 = 14.4Peach 7.0× 3.6 = 25.2Neopolitan 17.0× 3.6 = 61.2Other 6.0× 3.6 = 21.6


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Problem 30.18In the United States, approximately 45% of the population has blood typeO; 40% type A; 11% type B; and 4% type AB. Illustrate this distribution ofblood types with a pie chart.

Solution.First we find the measure of the central angle of each category:

Blood type O 45× 3.6 = 162Blood type A 40× 3.6 = 144Blood type B 11× 3.6 = 39.60Blood type AB 4× 3.6 = 14.40


Problem 30.19

Make a pictograph to represent the data in the following table. Use torepresent 10 glasses of lemonade.

Day FrequencyMonday 15Tuesday 20Wednesday 30Thursday 5Friday 10

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Solution.

Problem 30.20The following pictograph shows the approximate number of people who speakthe six common languages on earth.(a) About how many people speak Spanish?(b) About how many people speak English?(c) About how many more people speak Mandarin than Arabic?

Solution.(a) 225 million since there are 2 and a quarter symbols.(b) 375 million people since there are 33

4symbols.

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(c) 675 million people speak Mandarin and 125 million people speal Arabic.Thus, there are 550 million more people speaking Mandarin than Arabic

Problem 30.21Twenty people were surveyed about their favorite pets and the result is shownin the table below.

Pet FrequencyDog 2Cat 5Hamster 3

Make a pictograph for the following table of data. Let stand for 2 votes.

Solution.

Problem 30.22Coach Lewis kept track of the basketball team’s jumping records for a 10-yearperiod, as follows:

Year ’93 ’94 ’95 ’96 ’97 ’98 ’99 ’00 ’01 ’02Record(nearest in) 65 67 67 68 70 74 77 78 80 81


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Solution.

(a)

(b) This scatterpplot represents a positive corrolation. This means that thejumping record increases with the years

Problem 30.23The gas tank of a National Motors Titan holds 20 gallons of gas. The fol-lowing data are collected during a week.

Fuel in tank (gal.) 20 18 16 14 12 10Dist. traveled (mi) 0 75 157 229 306 379


Solution.(a)

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(b)The plot represents a positive corrolation. The more gas consumed thelarger the distance traveled

Problem 31.1Jenny averaged 70 on her quizzes during the first part of the quarter and 80on her quizzes during the second part of the quarter. When she found outthat her final average for the quarter was not 75, she went to argue with herteacher. Give a possible explanation for Jenny’s misunderstanding.

Solution.She could have taken a different number of quizzes during the first part ofthe quarter than the second part

Problem 31.2Suppose the following circle graphs are used to illustrate the fact that thenumber of elementary teaching majors at teachers’ colleges has doubled be-tween 1993 and 2003, while the percent of male elementary teaching majorshas stayed the same. What is misleading about the way the graphs are con-structed?

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Solution.When the radius of a circleiss doubled, the area is quadrupled, which ismisleading since the population has only doubled. Recall that the area of acircle of radius r is A = πr2

Problem 31.3What is wrong with the following line graph?

Solution.The horizontal axis does not have uniformly sized intervals and both thehorizontal axis and the graph are not labeled

Problem 31.4Doug’s Dog Food Company wanted to impress the public with the magnitudeof the company’s growth. Sales of Doug’s Dog Food had doubled from 2002to 2003, so the company displayed the following graph, in which the radiusof the base and the height of the 2003 can are double those of the 2002 can.What does the graph really show with respect to the growth of the company?

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Solution.The three-dimensional drawing distorts the graph. The result of doublingthe radius and the height of the can is to increase the volume by a factor of8. Recall that the volume of the can is V = πr2h where r is the radius of thecircle at the base and h is the height of the can

Problem 31.5What’s wrong with the following graph?

Solution.No label on the vertical axis, so we cannot compare actual sales. Also, thereis no scale on the vertical axis

Problem 31.6Refer to the following pictograph:

Ms McNulty claims that on the basis of this information, we can concludethat men are worse drivers than women. Discuss whether you can reach

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that conclusion from the pictograph or you need more information. If moreinformation is needed, what would you like to know?

Solution.One would need more information; for example, is teh graph in percentagesor actual numbers?

Problem 31.7Larry and Marc took the same courses last quarter. Each bet that he wouldreceive the better grades. Their courses and grades are as follows:

Course Larry’s Grades Marc’s GradesMath(4 credits) A CChemistry(4 credits) A CEnglish(3 credits) B BPsychology(3 credits) C ATennis(1 credit) C A

Marc claimed that the results constituted a tie, since both received 2 A’s,1 B, and 2 C’s. Larry said that he won the bet because he had the higherGPA for the quarter. Who is correct?(Allow 4 points for A, 3 points for B,2 points for C, 1 point for D, and 0 point for F.)

Solution.Let’s find the GPA of each.Larry’s GPA: 4×4+4×4+3×3+3×2+1×2

15≈ 3.27

Marc’s GPA: 4×2+4×2+3×3+3×4+4×115

≈ 2.73Hence, Larry is right and Marc’s is wrong

Problem 31.8Oil prices went up 20% one year and 30% the next. Is it true that over thetwo years, prices went up 50%?

Solution.Let x be the oil price two years ago. Then last year the price was 1.2x dollarsand this year the new price is 1.2x + (0.3)(1.2x) = 1.56x so that the pricewent up 56% over the past two years

Problem 31.9True or false? My rent went down 10% last year and then rose 20% this year.Over the two years my rent went up by 10%.

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Solution.Suppose that the rent was x dollars two years ago. Then the rent last yearwas x− 0.1x = 0.9x dollars. The rent this year is 0.9x + 0.2(0.9x) = 1.08x.Hence, the rent went up 8% over the two years

Problem 31.10Which graph could be used to indicate a greater decrease in the price ofgasoline? Explain.

Solution.Graph A. Graph B is misleading since in one year a van was used as a symboland in the next a small compact car was used. The same symbol should beused in a good pictograph

Problem 32.1Find (a) the mean, (b) median, and (c) the mode for the following collectionof data:

60 60 70 95 95 100

Solution.(a) x = 60+60+70+95+95+100

6= 80

(b) The median rank is MR = 6+12

= 3.5 so that the median is 70+952

= 82.5(c) The mode is 60 and 95

Problem 32.2Suppose a company employs 20 people. The president of the company earns$200,000, the vice president earns $75,000, and 18 employees earn $10,000each. Is the mean the best number to choose to represent the ”average”salary of the company?

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Solution.The mean salary for this company is

$200, 000 + $75, 000 + 18($10, 000)

20= $22, 750.

In this case, the mean salary of $ 22,750 is not representative. Either themedian or mode, both of which are $ 10,000, would describe the typical salarybetter

Problem 32.3Suppose nine students make the following scores on a test:

30, 35, 40, 40, 92, 92, 93, 98, 99

Is the median the best ”average” to represent the set of scores?

Solution.The median rank is MR = 9+1

2= 5 so that the median score is 92. From that

score one might infer that individuals all scored very well, yet 92 is certainlynot a typical score. In this case, the mean of approximately 69 might bemore appropriate than the mode

Problem 32.4Is the mode an appropriate ”average” for the following test scores?

40, 42, 50, 62, 63, 65, 98, 98

Solution.The mode is 98 because this score occurs most frequently. The score of 98 isnot representative of the set of data because of the large spread of scores.

Problem 32.5The 20 meetings of a square dance club were attended by 26, 25, 28, 23, 25,24, 24, 23, 26, 26, 28, 26, 24, 32, 25, 27, 24, 23, 24, and 22 of its members.Find the mode, median, and mean.

Solution.The mode is 24. The sum of the given 20 numbers is 483 so that the meanis 24.15. To find the median we order the numbers for smallest to largest toobtain

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22 23 23 23 24 24 24 24 24 25 25 25 26 26 26 26 27 28 28 32

The median rank is MR = 20+12

= 10.5 so that the median is 25

Problem 32.6If the mean annual salary paid to the top of three executives of a firm is$96, 000, can one of them receive an annual salary of $300, 000?

Solution.If we denote the salaries of the three executives by x1, x2, and x3 then x1 +x2 + x3 = 3($96, 000) = $288, 000. Thus, none can have a salary of $ 300,000

Problem 32.7An instructor counts the final examination in a course four times as much asof the four one-hour examinations. What is the average grade of a studentwho received grades of 74, 80, 61, and 77 in the four one- hour examinationsand 83 in the final examination?

Solution.The average mean is

74 + 80 + 61 + 77 + 4(83)

8= 78

Problem 32.8In 1980 a college paid its 52 instructors a mean salary of $13, 200, its 96assistant professors a mean salary of $15, 800, its 67 associate professors amean salary of $18, 900, and its 35 full professors a mean salary of $23, 500.What was the mean salary paid to all the teaching staff of this college?

Solution.The mean salary is

52($13, 200) + 96($15, 800) + 67($18, 900) + 35($23, 500)

52 + 96 + 67 + 35= $17, 168

Problem 32.9The following table gives the average costs of a single-lens reflex camera:

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800 650 300 430 560 470 640 830400 280 800 410 360 600 310 370

(a) Rank the data from smallest to largest.(b) Find the quartiles Q1, Q2, and Q3.(c) Make a box-and-whisker plot.

Solution.(a)

280 300 310 360 370 400 410 430470 560 600 640 650 800 800 830

(b) The median rank of the whole set of data is MR = 16+12

= 8.5. So thesecond quartile is the average of the 8th and 9th numbers of the list, i.e.,Q2 = 430+470

2= 450. This number split the set of data into two sets each with

8 numbers and with median rank MR = 8+12

= 4.5. Thus, Q1 = 360+3702

= 365and Q3 = 640+650

2= 645

(c)

Problem 32.10Mr. Eyha took a general aptitude test and scored in the 82nd percentile foraptitude in accounting. What percentage of the scores were at or below hisscore? What percentage were above?

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Solution.The 82nd percentile means that 82% of the scores were at or below his scoreand 18% were above his score

Problem 32.11At Center Hospital there is a concern about the high turnover of nurses. Asurvey was done to determine how long (in months) nurses had been in theircurrent positions. The responses of 20 nurses were:

23 2 5 14 25 36 27 42 12 87 23 29 26 28 11 20 31 8 36

(a) Rank the data.(b) Make a box-and-whisker plot of the data.(c) What are your conclusions from the plot?

Solution.(a)

2 5 7 8 8 11 12 14 20 2323 25 26 27 28 29 31 36 36 42

(b) The three quartiles are Q1 = 9.5, Q2 = 23, and Q3 = 28.5. The box-and-whisker plot is given next.

347

(c) Note that the median of the given set of data, i.e., Q2 = 23, is abovethe center of the box. This shows that there are more nurses who have beenin the job for more than 23 months compared to those who have been in thejob for less than 23 months

Problem 32.12The following are the wind velocities reported at 6 P.M.on six consecutivedays: 13,8,15,11,3 and 10. Find the range, sample mean, sample variance,and sample standard deviation.

Solution.The range is 15−3 = 12. The mean is 13+8+15+11+3+10

6= 60

6= 10. The sample

variance is

s2 =(13− 10)2 + (8− 10)2 + (15− 10)2 + (11− 10)2 + (10− 10)2

5= 17.6

and the sample standard deviation is s =√

17.6 ≈ 4.2

Problem 32.13An airline’s records show that the flights between two cities arrive on theaverage 4.6 minutes late with a standard deviation of 1.4 minutes. At leastwhat percentage of its flights between these two cities arrive anywhere be-tween 1.8 minutes late and 7.4 minutes late?

Solution.1.8 and 7.4 are within two standard deviations of the mean so according tothe emperical rule, 95% of the flights arrive anywhere between 1.8 minuteslate and 7.4 minutes late

Problem 32.14One patient’s blood pressure, measured daily over several weeks, averaged182 with a standard deviation of 5.3, while that of another patient averaged124 with a standard deviation of 9.4. Which patient’s blood pressure isrelatively more variable?

Solution.Since 5.3

182= 2.9% and 9.4

124= 7.6% then patient’s 2 blood pressure is more

variable

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Problem 32.15By sampling different landscapes in a national park over a 2-year period, thenumber of deer per square kilometer was determined. The results were (deerper square kilometer)

30 20 5 29 58 720 18 4 29 22 9

Compute the range, sample mean, sample variance, and sample standarddeviation.

Solution.The range is 58− 4 = 54. The mean is the sum of the given numbers dividedby 12, i.e. 251

12≈ 20.92. Using a calculator one finds the variance to be about

225 and the standard deviation to be about 15

Problem 32.16A researcher wants to find the number of pets per household. The researcherconducts a survey of 35 households. Find the sample variance and standarddeviation.

0 2 3 1 01 2 3 1 01 2 1 1 03 2 1 1 14 1 2 2 43 2 1 2 32 3 4 0 2

Solution.The mean is 61

35≈ 1.74. The variance is 1.38 and standard deviation is 1.17

Problem 32.17Suppose two machines produce nails which are on average 10 inches long. Asample of 11 nails is selected from each machine.

Machine A: 6, 8, 8, 10, 10, 10, 10, 10, 12, 12, 14.Machine B: 6, 6, 6, 8, 8, 10, 12, 12, 14, 14, 14.

Which machine is better than the other?

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Solution.The standard deviation of Machine A is about 2.19 whereas that of MachineB is 3.35. Hence, Machine A is better than Machine B since more nails areclose to the mean than the case of Machine B

Problem 32.18Find the missing age in the following set of four student ages.

Student Age Deviation from the MeanA 19 -4B 20 -3C ? 1D 29 6

Solution.Note that each time the age increases by 1 the corresponding deviation in-creases by 1. Thus, the missing age 24 since the change in deviation is1− (−4) = 5 so that 19 + 5 = 24

Problem 32.19The maximum heart rates achieved while performing a particular aerobicexercise routine are measured (in beats per minute) for 9 randomly selectedindividuals.

145 155 130 185 170 165 150 160 125

(a) Calculate the range of the time until failure.(b) Calculate the sample variance of the time until failure.(c) Calculate the sample standard variation of the time until failure.

Solution.(a) 185− 125 = 60(b) The sample variance is about 361.11(c) The sample standard variation is

√361.11 ≈ 19

Problem 32.20The following data gives the number of home runs that Babe Ruth hit ineach of his 15 years with the New York Yankees baseball team from 1920 to1934:

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54 59 35 41 46 25 47 60 54 46 49 46 41 34 22.

The following are the number of home runs that Roger Maris hit in each ofthe ten years he played in the major leagues from 1957 on:

8 13 14 16 23 26 28 33 39 61

Calculate the mean and standard deviation for each player’s data and com-ment on the consistency of performance of each player.

Solution.The mean of Babe Ruth is 659

15≈ 43.93 and that of Roger Maris is 261

10= 26.1

Ruth’s standard deviation is 11.25 and that of Maris is 15.61. Since 11.2543.93

≈25.6% and 15.61

26.1= 59.8% than Ruth is more consisten than Maris

Problem 32.21An office of Price Waterhouse Coopers LLP hired five accounting traineesthis year. Their monthly starting salaries were: $2536; $2173; $2448; $2121;and $2622.

(a) Compute the population mean.(b) Compute the population variance.(c) Compute the population standard deviation.

Solution.(a) 2536+2173+2448+2121+2622

5= $2380

(b) $ 49,363.50(c) $ 222.18

Problem 32.22On a final examination in Statistics, the mean was 72 and the standarddeviation was 15. Assuming normal distribution, determine the z-score ofstudents receiving the grades (a) 60, (b) 93, and (c) 72.

Solution.(a) z = 60−72

15= −0.8

(b) z = 93−7215

= 1.4(c) z = 72−72

15= 0

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Problem 32.23Referring to the previous exercise, find the grades corresponding to the z-score z = 1.6.

Solution.

x = 72 + 15(1.6) = 96

Problem 32.24If z1 = 0.8, z2 = −0.4 and the corresponding x-values are x1 = 88 andx2 = 64 then find the mean and the standard deviation, assuming we have anormal distribution.

Solution.We have 88 = µ + 0.8σ and 64 = µ− 0.4σ. Subtracting the second equationfrom the first we find 1.2σ = 24 so that σ = 24

1.2= 20. Replacing this value

in the first equation and solving for µ we find µ = 72

Problem 32.25A student has computed that it takes an average (mean) of 17 minutes witha standard deviation of 3 minutes to drive from home, park the car, and walkto an early morning class. Assuming normal distribution,

(a) One day it took the student 21 minutes to get to class. How manystandard deviations from the average is that?(b) Another day it took only 12 minutes for the student to get to class. Whatis this measurement in standard units?(c) Another day it took him 17 minutes to be in class. What is the z-score?

Solution.(a) since 21− 17 = 4 then 21 is 11

3standard deviation from the mean.

(b) Since 17 − 12 = 5 then 12 minutes is 123

standard deviation from themean.(c) z = 17−17

3= 0

Problem 32.26Mr. Eyha’s z-score on a college exam is 1.3. If the x-scores have a mean of480 and a standard deviation of 70 points, what is his x-score?

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Solution.

x = 480 + 1.3(70) = 571

Problem 32.27(a) If µ = 80, σ = 10, what is the z-score for a person with a score of 92?(b) If µ = 65, σ = 12, what is the raw score for a z-score of -1.5?

Solution.(a) z = 92−80

10= 1.2

(b) x = 65− 1.5(12) = 47

Problem 32.28Sketch a normal curve. Mark the axis corresponding to the parameter µ andthe axis corresponding to µ + σ and µ− σ.

Solution.

Problem 32.29For the population of Canadian high school students, suppose that the num-ber of hours of TV watched per week is normally distributed with a mean of20 hours and a standard deviation of 4 hours. Approximately, what percent-age of high school students watch

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(a) between 16 and 24 hours per week?(b) between 12 and 28 hours per week?(c) between 8 and 32 hours per week?

Solution.(a) Within one standard deviation: 68%(b) Within two standard deviations: 95%(c) Withoin three standard deviations: 99.7%

Problem 32.30The length of human pregnancies from conception to birth varies accord-ing to a distribution that is approximately normal with mean 266 days andstandard deviation 16 days. Use the emperical rule to answer the followingquestions.

(a) Between what values do the lengths of the middle 95% of all pregnanciesfall?(b) How short are the shortest 2.5% of all pregnancies?

Solution.(a) Between 266− 2(16) = 234 days and 266 + 2(16) = 298 days.(b) less than or equal to 234 days

Problem 33.1An experiment consists of flipping a fair coin twice and recording each flip.Determine its sample space.

Solution.

S = {HH,HT, TH, TT}

Problem 33.2Three coins are thrown. List the outcomes which belong to each of the fol-lowing events.

(a) exactly two tails (b) at least two tails (c) at most two tails.

Solution.(a) {TTH, HTT, THT}(b) {TTT, TTH, HTT, THT}(c) {TTH, HTT, THT, THH, HTH,HHT, HHH}

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Problem 33.3For each of the following events A, B, C, list and count the number of out-comes it contains and hence calculate the probability of A, B or C occurring.

(a) A = ”throwing 3 or higher with one die”,(b) B = ”throwing exactly two heads with three coins”,(c) C = ”throwing a total score of 14 with two dice”.

Solution.(a) A = {3, 4, 5, 6}. P (A) = 4

6= 2

3

(b) B = {TTH, HTT, THT}. P (B) = 38

(c) C = ∅. P (C) = 0

Problem 33.4An experiment consists of throwing two four-faced dice.

(a) Write down the sample space of this experiment.(b) If E is the event ’total score is at least 4’ list the outcomes belonging toEc.(c) If each die is fair find the probability that the total score is at least 6when the two dice are thrown. What is the probability that the total scoreis less than 6?(d) What is the probability that a double: (i.e. {(1, 1), (2, 2), (3, 3), (4, 4)})will not be thrown?(e) What is the probability that a double is not thrown nor is the scoregreater than 6?

Solution.(a) S = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2),(3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}(b) Ec = {(1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1),(4, 2), (4, 3), (4, 4)}(c) P ({(2, 4), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)}) = 6

16= 3

8. The probability

that the total score is less than 6 is 1− 38

= 58

(d) 1216

= 34

(e) P ({(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (3, 1), (3, 2), (4, 1)}) = 816

= 12

Problem 33.5A lot consists of 10 good articles, 4 with minor defects and 2 with major

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defects. One article is chosen at random. Find the probability that:

(a) it has no defects,(b) it has no major defects,(c) it is either good or has major defects.

Solution.(a) 4

10= 40%

(b) 810

= 45

(c) 610

= 35

Problem 33.6Consider the experiment of spinning the pointer on the game spinner pic-tured below. There are three possible outcomes, that is, when the pointerstops it must point to one of the three colors. (We rule out the possibility oflanding on the border between two colors.)

(a) What is the probability that the spinner is pointing to the red area?(b) What is the probability that the spinner is pointing to the blue area?(c) What is the probability that the spinner is pointing to the green area?

Solution.(a) 1

2

(b) 23

(c) 16

Problem 33.7Consider the experiment of flipping a coin three times. If we denote a

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head by H and a tail by T, we can list the 8 possible ordered outcomes as(H, H, H), (H, H, T ), · · · each of which occurs with probability of 1/8. Finishlisting the remaining members of the sample space. Calculate the probabilityof the following events:

(a) All three flips are heads.(b) Exactly two flips are heads.(c) The first flip is tail.(d) At least one flip is head.

Solution.(a) 1

8

(b) 38

(c) 48

= 12

(d) 78

Problem 33.8Suppose an experimet consists of drawing one slip of paper from a jar con-taining 12 slips of paper, each with a different month of the year written onit. Find each of the following:

(a) The sample space S of the experiment.(b) The event A consisting of the outcomes having a month begining with J.(c) The event B consisting of outcomes having the name of a month that hasexactly four letters.(d) The event C consisting of outcomes having a month that begins with Mor N.

Solution.(a) The sample space consists of the 12 months of the year.(b) A = {January, June, July}(c) B = {June, July}(d) C = {March, May, November}

Problem 33.9Let S = {1, 2, 3, · · · , 25}. If a number is chosen at random, that is, with thesame chance of being drawn as all other numbers in the set, calculate eachof the following probabilities:

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(a) The even A that an even number is drawn.(b) The event B that a number less than 10 and greater than 20 is drawn.(c) The event C that a number less than 26 is drawn.(d) The event D that a prime number is drawn.(e) The event E that a number both even and prime is drawn.

Solution.(a) A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24} so that P (A) = 12

25

(b) B = ∅ so that P (B) = 0(c) C = S so that P (C) = 1(d) D = {2, 3, 5, 7, 11, 13, 17, 19, 23} so that P (D) = 9

25

(e) E = {2} so that P (E) = 125

Problem 33.10Consider the experiment of drawing a single card from a standard deck ofcards and determine which of the following are sample spaces with equallylikely outcomes:(a) {face card, not face card}(b) {club, diamond, heart, spade}(c) {black, red}(d) {king, queen, jack, ace, even card, odd card}

Solution.(a) Sincce P ({face card}) = 12

52= 3

13and P ({not a face card}) = 40

52= 10

13

then the outcomes are not equally likely.(b) Since P (club) = P (diamond) = P (heart) = P (spade) = 1

4then the

outcomes are equally likely.(c) Since P (black) = P (red) = 1

2then the outcomes are equally likely.

(d) Since P (King) = 452

= 113

and P (even card) = 2052

= 513

then the outcomesare not equally likely

Problem 33.11An experiment consists of selecting the last digit of a telephone number.Assume that each of the 10 digits is equally likely to appear as a last digit.List each of the following:(a) The sample space(b) The event consisting of outcomes that the digit is less than 5

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(c) The event consisting of outcomes that the digit is odd(d) The event consisting of outcomes that the digit is not 2(e) Find the probability of each of the events in (b) - (d)

Solution.(a) S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}(b) A = {0, 1, 2, 3, 4}(c) B = {1, 3, 5, 7, 9}(d) C = {0, 1, 3, 4, 5, 6, 7, 8, 9}(e) P (A) = 5

10= 1

2, P (B) = 1

2, P (C) = 9

10

Problem 33.12Each letter of the alphabet is written on a seperate piece of paper and placedin a box and then one piece is drawn at random.(a) What is the probability that the selected piece of paper has a vowel writ-ten on it?(b) What is the probability that it has a consonant written on it?

Solution.(a) Since there are five vowels then the probability is 5

26

(b) 2126

Problem 33.13The following spinner is spun:

Find the probabilities of obtaining each of the following:(a) P(factor of 35)(b) P(multiple of 3)(c) P(even number)(d) P(11)(e) P(composite number)(f) P(neither prime nor composite)

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Solution.(a) The factors of 35 in the figure are 1, 5, and 7 so that the probability is 3

8

(b) The multiples of 3 in the figure are 3 and 6 so that the probability is28

= 14

(c) The outcomes of this event are 2,4,6,8 so that the probability is 48

= 12

(d) P (11) = 0(e) The composite numbers, i.e, not prime numbers, are 4,6,8 so that theprobability is 3

8

(f) The only number that is neither prime nor composite is 1 so that theprobability is 1

8

Problem 33.14An experiment consists of tossing four coins. List each of the following.

(a) The sample space(b) The event of a head on the first coin(c) The event of three heads

Solution.(a) S = {HHHH,HHHT,HHTH, HHTT,HTHH, HTHT,HTTH,HTTT, THHH,THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}(b) E = {HHHH,HHHT,HHTH, HHTT,HTHH, HTHT,HTTH,HTTT}(c) F = {HHHH,HHHT,HHTH, HTHH, THHH}

Problem 33.15Identify which of the following events are certain, impossible, or possible.(a) You throw a 2 on a die(b) A student in this class is less than 2 years old(c) Next week has only 5 days

Solution.(a) Possible(b) impossible(c) impossible

Problem 33.16Two dice are thrown. If each face is equally likely to turn up, find thefollowing probabilities.

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(a) The sum is even(b) The sum is not 10(c) The sum is a prime(d) The sum is less than 9(e) The sum is not less than 9

Solution.(a) E = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1),(5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} so P (E) = 18

36= 1

2

(b) If F is event that the sum is 10 then F = {(4, 6), (6, 4), (5, 5)} so thatthe probability that the sum is not 10 is 1− 3

36= 1− 1

12= 11

12

(c) G = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6),(6, 1), (6, 5)} so that P (G) = 15

36= 5

12

(d) H = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1),(3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2)}so that P (H) = 26

36= 13

18

(e) 1− 1318

= 518

Problem 33.17What is the probability of getting yellow on each of the following spinners?

Solution.(a) 3

8(b) 3

6= 1

2

Problem 33.18A department store’s records show that 782 of 920 women who entered thestore on a Saturday afternoon made at least one purchase. Estimate theprobability that a woman who enters the store on a Saturday afternoon willmake at least one purchase.

Solution.The estimated probability is 782

920= 85%

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Problem 33.19Suppose that a set of 10 rock samples includes 3 that contain gold nuggets.If you were to pick up a sample at random, what is the probability that itincludes a gold nugget?

Solution.The probability is 3

10= 30%

Problem 33.20When do creative people get their best ideas? A magazine did a survey of 414inventors (who hold U.S. patents) and obtained the following information:

Time of Day When Best Ideas Occur

Time Number of Inventors6 A.M. - 12 noon 4612 noon - 6 P.M. 1886 P.M. - 12 midnight 6312 midnight - 6 A.M. 117

Assuming that the time interval includes the left limit and all the times upto but not including the right limit, estimate the probability (to two decimalplaces) that an inventor has a best idea during the time interval 6 A.M. - 12noon.

Solution.The estimated probability is 46

414≈ 11.1%

Problem 33.21Which of the following are mutually exclusive? Explain your answers.

(a) A driver getting a ticket for speeding and a ticket for going througha red light.(b) Being foreign-born and being President of the United States.

Solution.(a) Not mutually exclusive since a driver can get a ticket for speeding andgoing through a red light.(b) Mutually exclusive since the President of the Unites States has to beborn in the US

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Problem 33.22If A and B are the events that a consumer testing service will rate a givenstereo system very good or good, P (A) = 0.22, P (B) = 0.35. Find(a) P (Ac);(b) P (A ∪B);(c) P (A ∩B).

Solution.(a) P (Ac) = 1− 0.22 = 0.78(b) P (A ∪B) = P (A) + P (B) = 0.22 + 0.35 = 0.57(c) P (A ∩B) = 0

Problem 33.23If the probabilities are 0.20, 0.15, and 0.03 that a student will get a failinggrade in Statistics, in English, or in both, what is the probability that thestudent will get a failing grade in at least one of these subjects?

Solution.

0.20 + 0.15− 0.03 = 0.32

Problem 33.24If A is the event ”drawing an ace” from a deck of cards and B is the event”drawing a spade”. Are A and B mutually exclusive? Find P (A ∪B).

Solution.No since there is a card that is spade and ace. P (A ∪B) = P (A) + P (B)−P (A ∩B) = 4

52+ 13

52− 1

52= 16

52= 4

13

Problem 33.25A bag contains 18 coloured marbles: 4 are coloured red, 8 are colouredyellow and 6 are coloured green. A marble is selected at random. What isthe probability that the ball chosen is either red or green?

Solution.

4

18+

6

18=

10

18=

5

9

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Problem 33.26Show that for any events A and B, P (A ∩B) ≥ P (A) + P (B)− 1.

Solution.Since P (A∩B) ≤ 1 then −P (A∩B) ≥ −1. Add P (A) + P (B) to both sidesto obtain P (A) + P (B)− P (A ∩ B) ≥ P (A) + P (B)− 1. But the left handside is just P (A ∪B).

Problem 33.27A golf bag contains 2 red tees, 4 blue tees, and 5 white tees.

(a) What is the probability of the event R that a tee drawn at randomis red?(b) What is the probability of the event ”not R” that is, that a tee drawn atrandom is not red?(c) What is the probability of the event that a tee drawn at random is eitherred or blue?

Solution.(a) Because the bag contains a total of 11 tees and 2 tees are red thenP (R) = 2

11

(b) P (not red) = 911

(c) Since the two events are mutually exclusive then P (R ∪ B) = P (R) +P (B) = 2

11+ 4

11= 6

11

Problem 33.28A fair pair of dice is rolled. Let E be the event of rolling a sum that is aneven number and P the event of rolling a sum that is a prime number. Findthe probability of rolling a sum that is even or prime?

Solution.Since E = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4),(4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} then P (E) = 18

36= 1

2. Since P =

{(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2),(3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)} then P (P ) = 15

36= 5

12. Also,

P (E ∩ P ) = 136

so that

P (E ∪ P ) = P (E) + P (P )− P (E ∩ P ) =18

36+

15

36− 1

36=

32

36=

8

9

364

Problem 33.29If events A nd B are from the same sample space, and if P(A)=0.8 andP(B)=0.9, can events A and B be mutually exclusive?

Solution.If A and B are mutually exclusive then P (A ∩ B) = 0 so that P (A ∪ B) =P (A) + P (B) = 0.8 + 0.9 = 1.7. But P (A∪B) is a probability and so it cannot exceed 1. Hence, A and B can not be mutually exclusive

Problem 34.1If each of the 10 digits is chosen at random, how many ways can you choosethe following numbers?

(a) A two-digit code number, repeated digits permitted.(b) A three-digit identification card number, for which the first digit cannotbe a 0.(c) A four-digit bicycle lock number, where no digit can be used twice.(d) A five-digit zip code number, with the first digit not zero.

Solution.(a) This is a decision with two steps. For the first step we have 10 choicesand for the second we have also 10 choices. By the Fundamental Principleof Counting there are 10 · 10 = 100 two-digit code numbers.(b) This is a decision with three steps. For the first step we have 9 choices,for the second we have 10 choices, and for the third we have 10 choices. Bythe Fundamental Principle of Counting there are 9 · 10 = 900 three-digitidentification card numbers.(c) This is a decision with four steps. For the first step we have 10 choices,for the second 9 choices, for the third 8 choices, and for fourth 7 choices. Bythe Fundamental Principle of Counting there are 10 · ·9 ·8 ·7 = 5040 four-digitcode numbers.(d) This is a decision with five steps. For the first step we have 9 choices, forthe second 10 choices, for the third 10 choices, for the fourth 10 choices, andfor the fifth 10 choices. By the Fundamental Principle of Counting there are9 · 10 · 10 · 10 · 10 = 90, 000 five-digit zip codes.

Problem 34.2(a) If eight horses are entered in a race and three finishing places are consid-ered, how many finishing orders can they finish?

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(b) If the top three horses are Lucky one, Lucky Two, and Lucky Three, inhow many possible orders can they finish?

Solution.(a) This is a decision with three steps. For the first finisher we have 8 choices,7 choices for the second and 6 choices for the third. By the FundamentalPrinciple of Counting there are 8 · 7 · 6 = 336 different first three finishers.(b) This is a decision with three steps. For the first finisher we have 3 choices,2 choices for the second and 1 choice for the third. By the FundamentalPrinciple of Counting there are 3 · 2 · 1 = 6 different arrangements.

Problem 34.3You are taking 3 shirts(red, blue, yellow) and 2 pairs of pants (tan, gray) ona trip. How many different choices of outfits do you have?

Solution.(a) This is a decision with two steps. For the choice of shirts there are 3choices and for the choice of pants there are 2 choices. By the FundamentalPrinciple of Counting there are 3 · 2 = 6 different choices of outfits

Problem 34.4The state of Maryland has automobile license plates consisting of 3 lettersfollowed by three digits. How many possible license plates are there?

Solution.This is a decision with 6 steps. For the first we have 26 choices, 26 choicesfor the second, 26 choices for the third and 10 for each of the fourth, fifthand sixth steps. By the Fundamental Principle of Counting we have: 26 ·26 ·26 · 10 · 10 · 10 = 17576000 different license plates

Problem 34.5A club has 10 members. In how many ways can the club choose a presidentand vice-president if everyone is eligible?

Solution.This is a decision with two steps. For the choice of president we have 10possible choices, for the choice of the vice-president we have 9 choices. Bythe Fundamental Principle of Counting there are 10 · 9 = 90 different ways

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Problem 34.6A lottery allows you to select a two-digit number. Each digit may be either1,2 or 3. Use a tree diagram to show the sample space and tell how manydifferent numbers can be selected.

Solution.

The different numbers are {11, 12, 13, 21, 22, 23, 31, 32, 33}

Problem 34.7In a medical study, patients are classified according to whether they haveblood type A, B, AB, or O, and also according to whether their blood pressureis low, normal, or high. Use a tree diagram to represent the various outcomesthat can occur.

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Solution.

Problem 34.8If a travel agency offers special weekend trips to 12 different cities, by air,rail, or bus, in how many different ways can such a trip be arranged?

Solution.This is a decision with two steps. The first is to choose the destination andwe have 12 choices. The second is to choose the way to travel and we havethree choices. By the Fundamental Principle of Counting there are 12·3 = 36ways for the trip

Problem 34.9If twenty paintings are entered in art show, in how many different ways canthe judges award a first prize and a second prize?

Solution.For the first prize we have 20 choices and for the second we have 19 choices.By the Fundamental Principle of Counting we have 20 · 19 = 380 possiblechoices

Problem 34.10In how many ways can the 52 members of a labor union choose a president,a vice-president, a secretary, and a treasurer?

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Solution.For the president we have 52 choices, for the vice-president we have 51 choices,for the secretary we have 50 choices, and for the treasurer we have 49 choices.By the Fundamental Principle of Counting there are 52·51·50·49 = 6, 497, 400possible choices

Problem 34.11Find the number of ways in which four of ten new movies can be ranked first,second, third, and fourth according to their attendance figures for the firstsix months.

Solution.For the first ranking we have 10 choices, for the second we have 9 choices,for the third 8 choices, and for the fourth 7 choices. By the FundamentalPrinciple of Counting there are 10 · 9 · 8 · 7 = 5040 possible ways

Problem 34.12To fill a number of vacancies, the personnel manager of a company has tochoose three secretaries from among ten applicants and two bookkeepersfrom among five applicants. In how many different ways can the personnelmanager fill the five vacancies?

Solution.This is a decision with two steps. For the choice of secretaries there areC(10, 3) different ways and for the choice of the bookkeepers there are C(5, 2).By the Fundamental Principle of Counting there are C(10, 3) · C(5, 2) =1200 different ways. The notation C(., .) will be introduced and discussed inSection 35

Problem 34.13A box contains three red balls and two blue balls. Two balls are to be drawnwithout replacement. Use a tree diagram to represent the various outcomesthat can occur. What is the probability of each outcome?

369

Solution.

Problem 34.14Repeat the previous exercise but this time replace the first ball before drawingthe second.

Solution.

Problem 34.15If a new-car buyer has the choice of four body styles, three engines, and tencolors, in how many different ways can s/he order one of these cars?

Solution.This is a decision with three stages. For the choice of body styles we have4 choices, for the engine three, and for the color ten. By the FundamentalPrinciple of Counting we have 4 · 3 · 10 = 120 different choices

370

Problem 34.16A jar contains three red gumballs and two green gumballs. An experimentconsists of drawing gumballs one at a time from the jar, without replacement,until a red one is obtained. Find the probability of the following events.

A: Only one draw is needed.B: Exactly two draws are needed.

C: Exactly three draws are needed.

Solution.(a) P (A) = 3

5

(b) P (B) = 25· 3

4= 3

10

(c) P (C) = 25· 1

4· 3

3= 1

10

Problem 34.17Consider a jar with three black marbles and one red marble. For the exper-iment of drawing two marbles with replacement, what is the probability ofdrawing a black marble and then a red marble in that order?


4· 1

4= 3

16

371

Problem 34.18A jar contains three marbles, two black and one red. Two marbles are drawnwith replacement. What is the probability that both marbles are black?Assume that the marbles are equally likely to be drawn.

Solution.The probability that both are black is 2

3· 2

3= 4

9

Problem 34.19A jar contains four marbles-one red, one green, one yellow, and one white.If two marbles are drawn without replacement from the jar, what is theprobability of getting a red marble and a white marble in that order?


4· 1

3= 1

12

Problem 34.20A jar contains 3 white balls and 2 red balls. A ball is drawn at random fromthe box and not replaced. Then a second ball is drawn from the box. Draw atree diagram for this experiment and find the probability that the two ballsare of different colors.


5· 2

4+ 2

5· 3

4= 3

5

372

Problem 34.21Suppose that a ball is drawn from the box in the previous problem, its colorrecorded, and then it is put back in the box. Draw a tree diagram for thisexperiment and find the probability that the two balls are of different colors.


5· 2

5+ 2

5· 3

5= 12

25

Problem 34.22Suppose there are 19 balls in an urn. They are identical except in color. 16of the balls are black and 3 are purple. You are instructed to draw out oneball, note its color, and set it aside. Then you are to draw out another balland note its color. What is the probability of drawing out a black on thefirst draw and a purple on the second?


19· 3

18= 8

57

Problem 34.23The row of Pascal’s triangle that starts 1, 4, · · · would be useful in findingprobabilities for an experiment of tossing four coins.(a) Interpret the meaning of each number.(b) Find the probability of exactly one head and three tails.(c) Find the probability of at least one tail turning up.

Solution.(a) There are 1 four-head outcome and 4 three-head outcomes.(b) From Pascals’ trinagle in the notes we find that the probability of exactlyone head and three tails is 4

16= 1

4

(c) 416

(one tail) + 616

(two tails) + 416

(three tails) + 116

(four tails) = 1516

373

Problem 34.24Four coins are tossed.

(a) Draw a tree diagram to represent the arrangements of heads (H) andtails (T).(b) How many outcomes involve all heads? three heads, one tail? two heads,two tails? one head, three tails? no heads?(c) How do these results relate to Pascal’s triangle?

Solution.(a)

(b) 1(4H0T), 4(3H1T), 6(2H2T), 4(1H3T), 1(0H4T)(c) These numbers are the entries of the fourth row in Pascal’s triangle

Problem 34.25A true-false problem has 6 questions.

(a) How many ways are there to answer the 6-question test?(b) What is the probability of getting at least 5 right by guessing the answersat random?

Solution.(a) This is a binary experiment. Each question is either true or false so bythe Principle of Counting there are 2 · 2 · 2 · 2 · 2 · 6 = 26 = 64 different ways.(b) 1

64(all six are right) + 6

64(five right and one wrong) = 7

64

Problem 34.26(a) Write the 7th row of Pascal’s triangle.(b) What is the probability of getting at least four heads when tossing sevencoins?

374

Solution.(a)1 7 21 35 35 21 7 1(b) 1

128+ 7

128+ 21

128+ 35

128= 64

128= 1

2

Problem 34.27Assume the probability is 1

2that a child born is a boy. What is the probability

that if a family is going to have four children, they will all be boys?

Solution.Using the fourth row of Pascal’s triangle with boy = head and girl = tail wefind 1

16

Problem 35.1Compute each of the following expressions.

(a) (2!)(3!)(4!)(b) (4× 3)!(c) 4 · 3!(d) 4!− 3!(e) 8!

5!

(f) 8!0!

Solution.(a) (2!)(3!)(4!) = 2 · 6 · 24 = 288(b) (4× 3)! = 12! = 479001600(c) 4 · 3! = 4 · 6 = 24(d) 4!− 3! = 24− 6 = 18(e) 8!

5!= 336

(f) 8!0!

= 8! = 40320

Problem 35.2Compute each of the following.

(a) P (7, 2) (b) P (8, 8) (c) P (25, 2)

Solution.(a) P (7, 2) = 7!

(7−2)!= 42

(b) P (8, 8) = 8!(8−8)!

= 8! = 40320

(c) P (25, 2) = 25!(25−2)!

= 25!23!

= 25 · 24 = 600

375

Problem 35.3Find m and n so that P (m, n) = 9!

6!

Solution.Since m!

(m−n)!= 9!

6!then m = 9 and n = 3

Problem 35.4How many four-letter code words can be formed using a standard 26-letteralphabet(a) if repetition is allowed?(b) if repetition is not allowed?

Solution.(a) 26 · 26 · 26 · 26 = 456976(b) P (26, 4) = 26!

(26−4)!= 26 · 25 · 24 · 23 = 358800

Problem 35.5Certain automobile license plates consist of a sequence of three letters fol-lowed by three digits.

(a) If no repetitions of letters are permitted, how many possible license platesare there?(b) If no letters and no digits are repeated, how many license plates arepossible?

Solution.(a) P (26, 3) · 10 · 10 · 10 = 15600000(b) P (26, 3) · P (10, 3) = 15600 · 720 = 11232000

Problem 35.6A combination lock has 40 numbers on it.

(a) How many different three-number combinations can be made?(b) How many different combinations are there if the numbers must be alldifferent?

Solution.(a) 40 · 40 · 40 = 64000(b) P (40, 3) = 40!

(40−3)!= 40!

37!= 59280

376

Problem 35.7(a) Miss Murphy wants to seat 12 of her students in a row for a class picture.How many different seating arrangements are there?(b) Seven of Miss Murphy’s students are girls and 5 are boys. In how manydifferent ways can she seat the 7 girls together on the left, and then the 5boys together on the right?

Solution.(a) P (12, 12) = 12! = 479001600(b) P (7, 7) · P (5, 5) = 7! · 5! = 5040 · 120 = 604800

Problem 35.8Using the digits 1, 3, 5, 7, and 9, with no repetitions of the digits, how many

(a) one-digit numbers can be made?(b) two-digit numbers can be made?(c) three-digit numbers can be made?(d) four-digit numbers can be made?

Solution.(a) 5(b) P (5, 2) = 20(c) P (5, 3) = 60(d) P (5, 4) = 120

Problem 35.9There are five members of the Math Club. In how many ways can thepositions of officers, a president and a treasurer, be chosen?

Solution.There are P (5, 3) = 20 different ways

Problem 35.10(a) A baseball team has nine players. Find the number of ways the managercan arrange the batting order.(b) Find the number of ways of choosing three initials from the alphabet ifnone of the letters can be repeated.

377

Solution.(a) P (9, 9) = 362880(b) P (26, 3) = 15600

Problem 35.11Compute each of the following: (a) C(7,2) (b) C(8,8) (c) C(25,2)

Solution.(a) C(7, 2) = 7!

2!(5−2)!= 7!

2!3!= 21

(b) C(8, 8) = 1(c) C(25, 2) = 300

Problem 35.12Find m and n so that C(m, n) = 13

Solution.Since C(m, n) = m!

n!(m−n)!= 13 then we can choose m = 13 and n = 1

Problem 35.13The Library of Science Book Club offers three books from a list of 42. Ifyou circle three choices from a list of 42 numbers on a postcard, how manypossible choices are there?

Solution.Since order is not of importance here then the number of different choices isC(42, 3) = 11480

Problem 35.14At the beginning of the second quarter of a mathematics class for elementaryschool teachers, each of the class’s 25 students shook hands with each of theother students exactly once. How many handshakes took place?

Solution.Since the handshake between persons A and B is the same as that betweenB and A, this is a problem of choosing combinations of 25 people two at atime. There are C(25, 2) = 300 different handshakes

Problem 35.15There are five members of the math club. In how many ways can the two-person Social Committee be chosen?

378

Solution.Since order is irrelevant than the different two-person committe is C(5, 2) =10

Problem 35.16A consumer group plans to select 2 televisions from a shipment of 8 to checkthe picture quality. In how many ways can they choose 2 televisions?

Solution.There are C(8, 2) = 28 different ways

Problem 35.17The Chess Club has six members. In how many ways(a) can all six members line up for a picture?(b) can they choose a president and a secretary?(c) can they choose three members to attend a regional tournament with noregard to order?

Solution.(a) P (6, 6) = 6! = 720 different ways(b) P (6, 2) = 30 ways(c) C(6, 3) = 20 different ways

Problem 35.18Find the smallest values m and n such that C(m, n) = P (15, 2)

Solution.We have C(m, n) = m!

n!(m−n)!= 15!

13!. It follows that the smallest values of m

and n are m = 15 and n = 0

Problem 35.19A school has 30 teachers. In how many ways can the school choose 3 peopleto attend a national meeting?


Problem 35.20Which is usually greater the number of combinations of a set of objects orthe number of permutations?

379

Solution.Recall that P (m, n) = m!

(m−n)!whereas C(m, n) = m!

n!(m−n)!. Since n! ≥ 1 then

1n!≤ 1. Multiply both sides by m!

(m−n)!to obtain m!

n!(m−n)!≤ m!

(m−n)!. Hence,

C(m,n) ≤ P (m, n)

Problem 35.21How many different 12-person juries can be chosen from a pool of 20 juries?


Problem 35.22John and Beth are hoping to be selected from their class of 30 as presidentand vice-president of the Social Committee. If the three-person committee(president, vice-president, and secretary) is selected at random, what is theprobability that John and Beth would be president and vice president?

Solution.There are 28 committees where John is the president and Beth is the vice-president. The number of groups of three out of 30 people is C(30, 3) = 4060.Hence, the probability of John being selected as president and Beth as vice-president is 28

4060≈ 0.0069

Problem 35.23There are 10 boys and 13 girls in Mr. Benson’s fourth-grade class and 12boys and 11 girls in Mr. Johnson fourth-grade class. A picnic committeeof six people is selected at random from the total group of students in bothclasses.

(a) What is the probability that all the committee members are girls?(b) What is the probability that the committee has three girls and threeboys?

Solution.(a) P (committee has six girls) = C(24,6)

C(46,6)≈ 0.014

(b) P (three boys and three girls) = C(22,3)·C(24,3)C(46,6)

≈ 0.333

380

Problem 35.24A school dance committee of 4 people is selected at random from a group of6 ninth graders, 11 eighth graders, and 10 seventh graders.

(a) What is the probability that the committee has all seventh graders?(b) What is the probability that the committee has no seventh graders?

Solution.(a) P (four seventh graders) = C(10,4)

C(27,4)≈ 0.012

(b) 1− 0.012 = 0.988

Problem 35.25In an effort to promote school spirit, Georgetown High School created IDnumbers with just the letters G, H, and S. If each letter is used exactly threetimes,

(a) how many nine-letter ID numbers can be generated?(b) what is the probability that a random ID number starts with GHS?

Solution.(a) This is a decision with three stages. The first stage is to find the numberof ways the three letters G can be arranged in 9 vacant positions. There areC(9, 3) ways. One these three positions were filled with three G, then weneed to arrange the three letters H in 6 positions. There are C(6, 3) differentways. Finally, there are C(3, 3) to arrange the three letters S in the threeremaining positions. By the Fundamental Principle of Counting there areC(9, 3) · C(6, 3) · C(3, 3) = 1680 different ID numbers.(b) The first three positions of the ID are occupied by GHS in order. Thereare 6 positions left to be used by the letters G, H and S repeated twice. Thereare C(6, 2) · C(4, 2) · C(2, 2) = 90. Thus, the probability that a random IDnumber starts with GHS is 90

1680= 3

56

Problem 35.26The license plates in the state of Utah consist of three letters followed bythree single-digit numbers.

(a) If Edward’s initials are EAM, what is the probability that his licenseplate will have his initials on it (in any order)?(b) What is the probability that his license plate will have his initials in thecorrect order?

381

Solution.(a) The total number of license plates consisting of three letters followed bythree digits is 26 · 26 · 26 · 10 · 10 · 10 = 17576000. The total number of plateswith the letters E,A, M in any order is 6 · 10 · 10 · 10 = 6000. Thus, theprobability that his license plate will have his initials on it (in any order) is

6000

17576000

(b) The probability in this case is 100017576000

Problem 36.1If the probability of a boy’s being born is 1

2, and a family plans to have four

children, what are the odds against having all boys?

Solution.The probability of having all four boys is 1

16. The odds against having all

four boys is 15 : 1 since1− 1

16116

=15

1

Problem 36.2If the odds against Deborah’s winning first prize in a chess tournament are3 to 5, what is the probability that she will win first prize?

Solution.The probability that she will win the first prize satisfies the equation

1− P (E)

P (E)=

3

5

Solving this equation for P (E) we find P (E) = 58

Problem 36.3What are the odds in favor of getting at least two heads if a fair coin is tossedthree times?

Solution.The probability of getting at least two heads is 1

8+ 3

8= 1

2. The odds in favor

is12

1− 12

= 1 or 1 : 1

382

Problem 36.4If the probability of rain for the day is 60%, what are the odds against itsraining?

Solution.The odds against rain is

1− 0.6

0.6=

4

6=

2

3or 2 : 3 qed

Problem 36.5On a tote board at a race track, the odds for Gameylegs are listed as 26:1.Tote boards list the odds that the horse will lose the race. If this is the case,what is the probability of Gameylegs’s winning the race?

Solution.The odds against winning is 26:1. The probability of winning the race satisfiesthe equation

1− P (E)

P (E)= 26

Solving for P (E) we find P (E) = 127

Problem 36.6If a die is tossed, what are the odds in favor of the following events?(a) Getting a 4(b) Getting a prime(c) Getting a number greater than 0(d) Getting a number greater than 6.

Solution.(a)

16

1− 16

= 15

or 5 to 1

(b) There are three prime numbers: 2, 3, and 5. The probability of get-ting a prime number is 1

2. The odds in favor is then

12

1− 12

= 1 or 1 : 1

(c) The probability of getting a number greater than 0 is 1. The odds infavor is 1 : 0(d) The probability of getting a number greater than 0 is 0 so that the oddin favor is 0 : 1

383

Problem 36.7Find the odds against E if P (E) = 3

4.

Solution.

1− P (E)

P (E)=

1− 34

34

=1

3or 1 : 3

Problem 36.8Find P(E) in each case.

(a) The odds in favor of E are 3:4(b) The odds against E are 7:3

Solution.(a) In this case P (E) satisfies P (E)

1−P (E)= 3

4. Solving for P (E) we find P (E) = 3

7

(b) Solving the equation 1−P (E)P (E)

= 73

for P (E) we find P (E) = 310

Problem 36.9Compute the expected value of the score when rolling two dice.

Solution.

Score 2 3 4 5 6 7 8 9 10 11 12# of outcome(s) 1 2 3 4 5 6 5 4 3 2 1

The expected value is

E = 2× 136

+ 3× 236

+ 4× 336

+ 5× 436

+ 6× 536

+ 7× 636

+ 8× 536

+ 9× 436

+ 10× 336

+ 11× 236

+ 12× 136

= 7

Problem 36.10A game consists of rolling two dice. You win the amounts shown for rollingthe score shown.

Score 2 3 4 5 6 7 8 9 10 11 12$ won 4 6 8 10 20 40 20 10 8 6 4

Compute the expected value of the game.

384

Solution.

E = 4× 136

+ 6× 236

+ 8× 336

+ 10× 436

+ 20× 536

+ 40× 636

+ 20× 536

+ 10× 436

+ 8× 336

+ 6× 236

+ 4× 136

= 503≈ $16.67

Problem 36.11Consider the spinner in Figure 36.1, with the payoff in each sector of thecircle. Should the owner of this spinner expect to make money over anextended period of time if the charge is $2.00 per spin?

Figure 36.1

Solution.

E = −1× 12

+ 0× 14

+ 1× 18

+ 2× 18

= −18

= −$0.125

So the owner will make on average 12.5 cents per spin

Problem 36.12You play a game in which two dice are rolled. If a sum of 7 appears, youwin $10; otherwise, you lose $2.00. If you intend to play this game for along time, should you expect to make money, lose money, or come out abouteven? Explain.

Solution.

E = 10× 1

6− 2× 5

6= 0

Therefore, you should come out about even if you play for a long time

Problem 36.13Suppose it costs $8 to roll a pair of dice. You get paid the sum of the numbersin dollars that appear on the dice. What is the expected value of this game?

385

Solution.

E = −6× 136− 5× 2

36− 4× 3

36− 3× 4

36− 2× 5

36

− 1× 636

+ 0× 536

+ 1× 436

+ 2× 336

+ 3× 236

+ 4× 136

= $− 1

Problem 36.14An insurance company will insure your dorm room against theft for a semester.Suppose the value of your possessions is $800. The probability of your beingrobbed of $400 worth of goods during a semester is 1

100, and the probability

of your being robbed of $800 worth of goods is 1400

. Assume that these arethe only possible kinds of robberies. How much should the insurance com-pany charge people like you to cover the money they pay out and to makean additional $20 profit per person on the average?

Solution.The expected value is

E = (400)× 1

100+ (800)× 1

400= $6

They should charge $ 6 + $ 20 = $ 26 for the policy to make an average gainof $ 20 per policy

Problem 36.15Consider a lottery game in which 7 out of 10 people lose, 1 out of 10 wins$50, and 2 out of 10 wins $35. If you played 10 times, about how much wouldyou expect to win?

Solution.

E = −1× 7

10+ 50× 1

10+ 35× 2

10=

113

10

So if you play this game 10 times you expect to win on average $ 113

Problem 36.16Suppose a lottery game allows you to select a 2-digit number. Each digitmay be either 1, 2, 3, 4, or 5. If you pick the winning number, you win $10.Otherwise, you win nothing. What is the expected payoff?

386

Solution.There are 5× 5 = 25 2-digit number. The expected value is

E = −1× 24

25+ 10× 1

25= −14

25

So if you play this game 25 times you expect to lose $ 14

Problem 36.17Suppose that A is the event of rolling a sum of 7 with two fair dice. Makeup an event B so that(a) A and B are independent.(b) A and B are dependent.

Solution.(a) Let B be the event ” It rains tomorrow”. Then A and B are independent.(b) Let B be the event ”of rolling an even sum”. Then A and B aredependent

Problem 36.18When tossing three fair coins, what is the probability of getting two tailsgiven that the first coin came up heads?

Solution.Let A be the event of getting two tails and B the event that the first coingcame up head. Then A = {TTT, TTH, THT,HTT}, B = {HHH, HTT, HHT, HTH}, A∩B = {HTT}. Thus, P (A) = P (B) = 1

2and P (A ∩ B)1

8. Finally, P (A|B) =

P (A∩B)P (B)

=1812

= 14

Problem 36.19Suppose a 20-sided die has the following numerals on its face:1, 1, 2, 2, 2, 3,3, 4, 5, 6, 7, ,8 9, 10, 11, 12, 13, 14, 15, 16. The die is rolled once and thenumber on the top face is recorded. Let A be the event the number is prime,and B be the event the number is odd. Find P (A|B) and P (B|A).

Solution.A = {2, 2, 2, 3, 3, 5, 7, 11, 13} and B = {1, 3, 3, 5, 7, 9, 11, 13, 15}. Thus, A ∩B = {3, 3, 5, 7, 11, 13}. Hence, P (A) = 9

20, P (B) = 9

20< and P (A ∩ B) = 6

20.

So

P (A|B) =P (A ∩B)

P (B)=

620920

=2

3

387

and

P (B|A) =P (A ∩B)

P (A)=

620920

=2

3

Problem 36.20What is the probability of rolling a 6 on a fair die if you know that the rollis an even number?

Solution.Let A be the event of rolling a 6 so that P (A) = 1

6. Let B = {2, 4, 6}. Then

P (B) = 12. Also, A ∩B = {6} so that P (A ∩B) = 1

6. Hence,

P (A|B) =P (A ∩B)

P (B)

1612

=1

3

Problem 36.21A red die and a green die are rolled. What is the probability of obtaining aneven number on the red die and a multiple of 3 on the green die?

Solution.Let A be the event of obtaining an even number on the red die and B theevent of obtaining a multiple of 3 on the green die. The two events areindepndent so that P (A ∩B) = P (A) · P (B) = 1

2· 1

3= 1

6

Problem 36.22Two coins are tossed. What is the probability of obtaining a head on thefirst coin and a tail on the second coin?

Solution.Let A be the outcome of getting a head on the first toss and B be the eventof getting a tail on the second toss. Then A and B are independent. Thus,P (A ∩B) = P (A) · P (B) = 1

2· 1

2= 1

4

Problem 36.23Consider two boxes: Box 1 contains 2 white and 2 black balls, and box2 contains 2 white balls and three black balls. What is the probability ofdrawing a black ball from each box?

388

Solution.Let A be the event of drawing a black ball from Box 1 and B that of Box 2.Then A and B are independent so that P (A∩B) = P (A) ·P (B) = 1

2· 3

5= 3

10

Problem 36.24A container holds three red balls and five blue balls. One ball is drawn anddiscarded. Then a second ball is drawn.(a) What is the probability that the second ball drawn is red if you drew ared ball the first time?(b) What is the probability of drawing a blue ball second if the first ball wasred?(c) What is the probability of drawing a blue ball second if the first ball wasblue?

Solution.(a) 2

7

(b) 57

(c) 47

Problem 36.25Consider the following events.

A: rain tomorrowB: You carry an umbrellaC: coin flipped tomorrow lands on heads

Which of two events are dependent and which are independent?

Solution.A and B are depedent. A and C, B and C are independent

Problem 36.26You roll a regular red die and a regular green die. Consider the followingevents.

A: a 4 on the red dieB: a 3 on the green dieC: a sum of 9 on the two dice

389

Tell whether each pair of events is independent or dependent.(a) A and B (b) B and C

Solution.(a) Independent(b) Depedent since a 3 on the green die makes it more likely 1

6that you will

get a sum of 9

Problem 37.1Find three objects in your classroom with surfaces that suggests commongeometric figures.

Solution.Answer may vary. A door (rectangle), a ceiling (a square), a wall clock (acircle)

Problem 37.2A fifth grader says a square is not a rectangle because a square has fourcongruent sides and rectangles don’t have that. A second fifth grader saysa square is a type of rectangle because it is a parallelogram and it has fourright angles.

(a) Which child is right?(b) How can you use the definitions to help the other child understand?

Solution.(a) The second child(b) Write the definition of a rectangle and show the child that a squaresatisfies it

Problem 37.3Suppose P={parallelograms}, Rh={rhombus}, S={squares}, Re={rectangles},T={trapezoids}, and Q={quadrilaterals}. Find(a) Rh ∩Re (b) T ∩ P

Solution.(a) Rh ∩Re = S(b) T ∩ P = ∅

390

Problem 37.4Organize the sets P, Rh, S, Re, T, and Q using Venn diagram.

Solution.

Problem 37.5(a) True or false? No scalene triangle is isosceles.(b) What shape is the diamond in a deck of cards?

Solution.(a) True. A scalene triangle is a triangle with the three sides of differentlengths. An isosceless triangle has two sides of equal length.(b) Rhombus

Problem 37.6How many squares are in the following design?

Solution.Nine square

Problem 37.7Tell whether each of the following shapes must, can, or cannot have at least

391

one right angle.


Solution.(a) Cannot(b) must(c) can(d) must(e) can

Problem 37.8In which of the following shapes are both pairs of opposite sides parallel?


Solution.(a), (b), (d), and (e)

Problem 37.9A square is also which of the following?(a) Quadrilateral(b) Parallelogram(c) Rhombus(d) Rectangle

Solution.(a), (b), (c), and (d)

Problem 37.10Fill in the blank with ”All”, ”Some”, or ”No”

392

(a) rectangles are squares.(b) parallelograms are trapezoids.(c) rhombuses are quadrilaterals.

Solution.(a) Some rectangles are squares.(b) No parallelograms are trapezoids.(c) All rhombuses are quadrilaterals.

Problem 37.11How many triangles are in the following design?

Solution.Nine triangles

Problem 37.12How many squares are found in the following figure?

Solution.Eighteen squares

Problem 37.13Given are a variety of triangles. Sides with the same length are indicated.Right angles are indicated.

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(a) Name the triangles that are scalene.(b) Name the triangles that are isosceles.(c) Name the triangles that are equilateral.(d) Name the triangles that contain a right angle.

Solution.(a) (b) and (d)(b) (a), (c), (e), and (f)(c) (a), (e)(d) (d) and (f)

Problem 37.14(a) How many triangles are in the figure?(b) How many parallelograms are in the figure?(c) How many trapezoids are in the figure?

Solution.(a) Thirteen triangles(b) Ten parallelograms(c) Nine Trapezoids

Problem 37.15If possible, sketch two parallelograms that intersect at exactly

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(a) one point(b) two points(c) three points(d) four points.

Solution.

Problem 37.16If possible, draw a triangle and a quadrilateral that intersect at exactly(a) one point(b) two points(c) three points.

Solution.

Problem 37.17Suppose P={parallelograms}, S={squares}, T={trapezoids}, and Q={quadrilaterals}.Find

(a) P ∩ S(b) P ∪Q

Solution.(a) P ∩ S = S(b) P ∪Q = Q

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Problem 37.18A fifth grader does not think that a rectangle is a type of parallelogram. Tellwhy it is.

Solution.A rectangle is a parallelogram since the opposite sides are parallel

Problem 37.19Tell whether each definition has sufficient information. If it is not sufficient,tell what information is missing.

(a) A rhombus is a quadrilateral with both pairs of opposite sides paral-lel.(b) A square is a quadrilateral with four congruent sides.(c) A rhombus is a quadrilateral that has four congruent sides.

Solution.(a) All four sides are congruent(b) All four angles are right angles(c) No additional information is needed

Problem 37.20Name properties that a square, parallelogram, and rhombus have in common.

Solution.At this point the only common property we can state is that the oppositesides are parallel

Problem 37.21How many different line segments are contained in the following portion of aline?

Solution.Sixteen different line segments

Problem 38.1Using Figure 38.5 show that m(∠1) + m(∠5) = 180◦.

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Solution.Clearly m(∠6) + m(∠5) = 180◦. But the angles ∠1 and ∠6 are alternateinterior angles so that m(∠6) = m(∠1). Thus, m(∠1) + m(∠5) = 180◦

Problem 38.2(a) How many angles are shown in the following figure?

(b) How many are obtuse?(c) How many are acute?

Solution.(a) Ten angles(b) ∠BOE, ∠AOE, ∠AOD(c) ∠DOE, ∠COD, ∠BOC, ∠AOB, ∠COE

Problem 38.3Find the missing angle in the following triangle.

Solution.The measure of the missing angle is 30◦

Problem 38.4In the figure below, m(∠BFC) = 55◦, m(∠AFD) = 150◦, m(∠BFE) =120◦. Determine the measures of m(∠AFB) and m(∠CFD).

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Solution.The angles ∠AFB and ∠BFE are complementary so that m(∠AFB) =180◦ −m(∠BFE) = 180◦ − 120◦ = 60◦. On the other hand, m(∠CFD) =150◦ −m(∠AFB)−m(∠BFC) = 150◦ − 60◦ − 55◦ = 35◦

Problem 38.5In the following figure m(∠1) = m(∠2)

2− 9◦. Determine m(∠1) and m(∠2).

Solution.Since m(∠1) + m(∠2) = 180◦ and m(∠1) = m(∠2)

2− 9◦ then m(∠2)

2− 9◦ +

m(∠2) = 180◦. Solving for 3m(∠2) = 2 × 189◦ = 378◦ or m(∠2) = 378◦

3=

126◦. Finally, m(∠1) = m(∠2)2

− 9◦ = 63◦ − 9◦ = 54◦

Problem 38.6Angles 3 and 8; 2 and 7 in Figure 38.5 are called alternate exterior angles.Show that m(∠3) = m(∠8).

Solution.We have m(∠3) = m(∠6) (corresponding angles) and m(∠6) = m(∠8) (ver-tical angles). Hence, m(∠3) = m(∠8)

Problem 38.7Following are the measures of ∠A, ∠B, ∠C. Can a triangle ∆ABC be madethat has the given angles? Explain.

(a) m(∠A) = 36◦, m(∠B) = 78◦, m(∠C) = 66◦.(b) m(∠A) = 124◦, m(∠B) = 56◦, m(∠C) = 20◦.(c) m(∠A) = 90◦, m(∠B) = 74◦, m(∠C) = 18◦.

Solution.(a) Since m(∠A) + m(∠B) + m(∠C) = 180◦ then ∆ABC can exist.(b) Since m(∠A) + m(∠B) + m(∠C) = 200◦ > 180◦ then ∆ABC can notexist.(c) Since Since m(∠A) + m(∠B) + m(∠C) = 182◦ > 180◦ then ∆ABC cannot exist

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Problem 38.8In the following figure AO is perpendicular to CO. If m(∠AOD) = 165◦ andm(∠BOD) = 82◦, determine the measures of ∠AOB and ∠BOC.

Solution.We have m(∠AOB) = 165◦−82◦ = 83◦. Also, m(∠BOC) = 90◦−m(∠AOB) =90◦ − 83◦ = 7◦

Problem 38.9In the figure below, find the measures of ∠1, ∠2, ∠3, and ∠4.

Solution.We have: 2x − 18◦ + x + 12◦ = 180◦ (See Problem 38.1). Solving for x wefind x = 186

3= 62◦. Now, m(∠1) = m(∠2) = 62◦− 15◦ = 47◦. Also, m(∠3) =

180◦−(62◦+12◦) = 106◦. Finally, m(∠4) = 180◦−m(∠2) = 180◦−47◦ = 133◦

Problem 38.10(a) How is a line segment different from a line?(b) What is the vertex of the angle ∠PAT?(c) How is AB different from AB?

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Solution.(a) A line segment has two endpoints; a line has none.(b) A(c) AB is the notation for the line segment with endpoints A and B whereasAB is its length

Problem 38.11A fourth grader thinks that m(∠A) is greater than m(∠B).

(a) Why might the child think this?(b) How could you put the angles together to show that m(∠A) < m(∠B)?

Solution.(a) ∠A appears to have longer sides.(b) Put the vertex of ∠B on top of the vertex of ∠A then you see that theopening of ∠A is smaller than that of ∠B

Problem 38.12In the figure below

(a) name two supplementary angles(b) name two complementary angles.

Solution.(a) ∠CED and ∠BEC(b) ∠CED and ∠AEC

Problem 38.13An angle measures 20◦. What is the measure of(a) its supplement? (b) its complement?

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Solution.(a) 160◦

(b) 70◦

Problem 38.14True or false? If false give an example.

(a) All right angles are congruent.(b) Two complementary angles are congruent.(c) Two supplementary angles are congruent.

Solution.(a) True(b) False. For example, 20◦ and 70◦ are complementary angles but are notcongruent.(c) False. For example, 20◦ and 160◦ are supplementary angles but are notcongruent

Problem 38.15Find the measures of the angles in the following figure.

Solution.m(∠4) = 35◦ (corresponding angles).m(∠6) = 180◦ − 35◦ = 145◦

m(∠2) = m(∠4) = 35◦ (vertical angles)m(∠8) = m(∠6) = 145◦ (vertical angles)m(∠7) = 35◦ (vertical angles)m(∠3) = m(∠1) = 180◦ −m(∠4) = 180◦ − 35◦ = 145◦

Problem 38.16How many degrees does the minute hand of a clock turn through

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(a) in sixty minutes?(b) in five minutes?(c) in one minute?

Solution.(a) 360◦

(b) 360◦

12= 30◦

(c) 6◦

Problem 38.17How many degrees does the hour hand of a clock turn through

(a) in sixty minutes?(b) in five minutes?

Solution.(a) 360◦

12= 30◦

(b) 5×30◦

60= 2.5◦

Problem 38.18Find the angle formed by the minute and hour hands of a clock at thesetimes.(a) 3:00 (b) 6:00 (c) 4:30 (d) 10:20

Solution.(a) 90◦

(b) 180◦

(c) The minute hand is on 6 and the hour hand is midway between 4 and 5.So the angle between the two is 30◦ + 15◦ = 45◦

(d) The minute hand is on 4 and the hour hand is 13

away from 10 so 23

awayfrom 11. Hence, the angle is 10◦ + 10◦ + 4× 30◦ = 140◦

Problem 38.19Determine the measures of the interior angles.

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Solution.

Problem 38.20(a) Can a triangle have two obtuse angles? Why?(b) Can a triangle have two right angles? Why?(c) Can a triangle have two acute angles?

Solution.(a) No, because an obtuse angle has measure greater than 90◦ and addingtwo such measures would exceed 180◦ which is the sum of all three interiorangles for any triangle.(b) No, same reaon as in (a).(c) Yes

Problem 38.21A hiker started heading due north, then turned to the right 38◦, then turnedto the left 57◦, and next turned right 9◦. To resume heading north, what turnmust be made?

Solution.From the figure below we see that m(∠4) + 9◦ = m(∠3) (vertical angles).Also, we have m(∠1) = 38◦ (alternate interior angles) and m(∠2) = 180◦ −57◦ = 123◦. Hence, m(∠3) = 180◦ − 38◦ − 123◦ = 19◦. It follows thatm(∠4) = 10◦. So the hiker needs to make a turn to the right by 10◦

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Problem 39.1Draw all lines of symmetry for the figure below.

Solution.

Problem 39.2(a) Find the vertical line(s) of symmetry of the letters A, U, V, T, Y.(b) Find the horizontal line(s) of symmetry of the letters D, E, C, B.(c) Find the vertical and horizontal line(s) of symmetry of the letters H, I,O, X.

404

Solution.


405

Solution.

Problem 39.4A regular polygon is a closed figure with all sides congruent. Find all thelines of symmetry for these regular polygons. Generalize a rule about thenumber of lines of symmetry for regular polygons.

Solution.

Problem 39.5Find the number of rotations of the following geometric shape.

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Solution.The rotational symmetries are: 72◦, 144◦, 216◦, 288◦ and 360◦


Solution.

Problem 39.7Let ABCD be a parallelogram.(a) Prove that ∠A and ∠B are supplementary.(b) Prove that angles ∠A and ∠C are congruent.

407

Solution.(a) According to the figure below we have that m(∠1) = m(∠B) (correspond-ing angles). But m(∠1) + m(∠A) = 180◦. Thus, m(∠A) + m(∠B) = 180◦

(b) From (a) we deduce that m(∠A)+m(∠B) = 180◦ and m(∠C)+m(∠B) =180◦. Thus, m(∠A) + m(∠B) = m(∠C) + m(∠B). Subtracting m(∠B) fromboth sides we find m(∠A) = m(∠C)

Problem 39.8Using the figure below find the height of the trapezoid(i.e., the distancebetween the parallel sides) in terms of a and b.

Solution.The triangle ABC is isosceles with BC = b−a

2. Thus, AB = b−a

2

Problem 39.9”The diagonals of a rectangle are congruent.” Why does this statement implythat the diagonals of a square must also be congruent?

Solution.Becuase is square is at the same time a rectangle

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Problem 39.10You learn the theorem that the diagonals of a parallelogram bisect each other.What other quadrilaterals must have this property?

Solution.Since a rhombus, a rectangle, and a square are all quadrilaterals so theyshare all the properties of a parallelogram

Problem 39.11Fill in the blank to describe the following circle with center N. Circle N isthe set of in a plane that are from .

Solution.Circle N is the set of points in a plane that are 8 in. from N

Problem 39.12A chord is a line segment with endpoints on a circle. True or false? Everychord is also a diameter of the circle.

Solution.False. The figure below shows a chord that is not a diameter

Problem 39.13The diameter of a circle divides its interior into two congruent regions. Howcan someone use this property in dividing a circular pizza or pie in half?

Solution.By cutting along a straight line that passes through the center of the pizaayou divide it in half

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Problem 39.14If possible draw a triangle and a circle that intersect at exactly(a) one point (b) two points (c) three points (d) four points

Solution.

Problem 39.15If possible draw two parallelograms that intersect at exactly(a) one point (b) two points (c) three points (d) four points

Solution.

Problem 39.16A quadrilateral has two right angles. What can you deduce about the mea-sures of the other two angles?

Solution.The remaining the angles are supplementary since the some of the interiorangles of a quadrilateral is always 360◦

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Problem 40.1List the numerical values of the shapes that are convex .

Solution.The only convex shapes are 1 and 3. The rest are concave since one can findtwo points with line segment not entirely contained in the shape

Problem 40.2Determine how many diagonals each of the following has:(a) 20-gon (b) 100-gon (c) n-gon

Solution.We look for a pattern. Let D denote the number of diagonals and n thenumber of sides. We have the following table

n 4 5 6 7D 2 5 9 14

One finds that the formula for D is D = n(n−3)2

.

(a) For n = 20, D = 20(20−3)2

= 170

(b) D = 100(100−3)2

= 4650

(v) D = n(n−3)2

Problem 40.3In a regular polygon, the measure of each interior angle is 162◦. How manysides does the polygon have?

Solution.Since 180◦ − 360◦

n= 162◦ then

180◦ − 360◦

n− 180◦ = 162◦ −−180◦

−360◦

n= −18◦

n = 360◦

18◦= 20 sides

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Problem 40.4Two sides of a regular octagon are extended as shown in the following figure.Find the measure of ∠1.

Solution.Consider the triangle ABC shown in the figure below. The exterior angleat A of the triangle is just an interior angle of the regular octagon and hasmeasure

180◦ − 360◦

n= 180◦ − 360◦

8= 135◦

Thus, m(∠CAB) = 45◦. Similarly, m(∠ABC) = 45◦. Hence, m(∠1) = 90◦

Problem 40.5Draw a quadrilateral that is not convex.

Solution.

Problem 40.6What is the sum of the interior angle measures of a 40-gon?

Solution.For an n−gone we have seen in the notes that the sum of all interior anglesis given by the formula

(n− 2)× 180◦

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For n = 40 we find(40− 2)× 180◦ = 6840◦

Problem 40.7A Canadian nickel has the shape of a regular dodecagon (12 sides). Howmany degrees are in each interior angle?

Solution.Each interior angle measures

180◦ − 360◦

12= 150◦

Problem 40.8Is a rectangle a regular polygon? Why or why not?

Solution.A rectangle is polygon that is not regular since the sides are not all congru-ent

Problem 40.9Find the measures of the interior, exterior, and central angles of a 12-gon.

Solution.The measure of the interior angle is

180◦ − 360◦

12= 150◦

The measure of the exterior angle which is also the measure of the centralangle is

360◦

12= 30◦

Problem 40.10Suppose that the measure of the interior angle of a regular polygon is 176◦.What is the measure of the central angle?

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Solution.Since the measure of the interior angle is given by the formula

180◦ − 360◦

n

and that of the central angle is 360◦

nthen from the previous equation we find

360◦

n= 180◦ − 176◦ = 4◦

Problem 40.11The measure of the exterior angle of a regular polygon is 10◦. How manysides does this polygon have?

Solution.Since the measure of the exterior angle is equal to the measure of the interiorangle then we have

360◦

n= 10◦

Solving this equation for n we find n = 360◦

10◦= 36 sides

Problem 40.12The measure of the central angle of a regular polygon is 12◦. How many sidesdoes this polygon have?

Solution.We have

360◦

n= 12◦

Solving this equation for n we find n = 360◦

12◦= 30 sides

Problem 40.13The sum of the measures of the interior angles of a regular polygon is 2880◦.How many sides does the polygon have?

Solution.The sum of the measures of the interior angles of a regular polygon is givenby the formula

(n− 2) · 180◦

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Thus,(n− 2) · 180◦ = 2880◦

n− 2 = 2880◦

180◦= 16

n− 2 + 2 = 16 + 2n = 18 sides

Problem 40.14How many lines of symmetry does each of the following have?

(a) a regular pentagon(b) a regular octagon(c) a regular hexagon.

Solution.

Problem 40.15How many rotational symmetry does a pentagon have?

Solution.Since each central angle of a pentagon measures

360◦

5= 72◦

then the rotational symmetries are: 72◦, 144◦, 216◦, 288◦ and 360◦

Problem 41.1Determine the measures of all dihedral angles of a right prism whose basesare regular octagons.

Solution.Since the prism is a right prism, the dihedral angles made by either base to

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any lateral face measure 90◦. Now, the measure of the interior angles of theregular octagon is

180◦ − 360◦

8= 135◦

then the lateral faces meet at 135◦. The dihedral angle between a lateral faceand a base is 90◦

Problem 41.2Consider the cube given in the figure below.

(a) How many planes are determined by the faces of the cube?(b) Which edges of the cube are parallel to edge AB?(c) Which edges of the cube are contained in lines that are skew to the linegoing through A and B?

Solution.(a) Six planes: ABCD, EFGH, ABFE, CDHG, BCGF, ADHE(b) CD, GH, EF(c) DH, CG, FG, EH

Problem 41.3If a pyramid has an octagon for a base, how many lateral faces does it have?

Solution.Eight lateral faces since a lateral face is determined by the apex and a sideof the octagon

Problem 41.4Determine the number of faces, vertices, and edges for a hexagonal pyramid.

Solution.There are seven faces, seven vertices, and twelve edges

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Problem 41.5Sketch drawings to illustrate different possible intersections of a square pyra-mid and a plane.

Solution.Some of the cross sections that can be obtained: A plane can intersect thepyramid at exactly one point. If the plane is perpendicular to the line con-necting the apex and the center of the square then a cross section is a square.If the plane crosses the base then the cross section is a rectangle. A cross-section can consists of a line segement, an edge of the pyramid. Finally, across-section can be a triangle

Problem 41.6Sketch each of the following.

(a) A plane and a cone that intersect in a circle.(b) A plane and a cylinder that intersect in a segment.(c) Two pyramids that intersect in a triangle.

Solution.

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Problem 41.7For the following figure, find the number of faces, edges, and vertices.

Solution.Nine faces, sixteen edges, Nine vertices

Problem 41.8For each of the following figures, draw all possible intersections of a planewith(a) Cube (b) Cylinder

Solution.

(b) A cross-section can be a point, a circle, a line segement, or an oval

Problem 41.9A diagonal of a prism is any segment determined by two vertices that do notlie in the same face. How many diagonals does a pentagonal prism have?

Solution.2× 5 = 10 diagonals

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Problem 41.10For each of the following, draw a prism and a pyramid that have the givenregion as a base:

(a) Triangle(b) Pentagon(c) Regular hexagon

Solution.

Problem 41.11If possible, sketch each of the following:(a) An oblique square prism(b) An oblique square pyramid

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(c) A noncircular right cone(d) A noncircular cone that is not right

Solution.

Problem 41.12A simple relationship among the number of faces(F), the number of edges(E),and the number of vertices(V) of any convex polyhedron exists and is knownas Euler’s formula. Use the following table to find this relationship.


Solution.

V + F = E + 2

where V=vertex, F=face, E=edge

Problem 41.13What is the sum of the angles at the each vertex of a(a) Tetrahedron(b) Octahedron(c) Cube(d) Icosahedron(e) Dodecahedron

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Solution.(a) The faces consist of equilateral triangles so that each interior angle mea-sures 60◦. Thus, the sum of the angles at each vertex is 180◦

(b) There are 8 faces consisting of equilateral triangles. Thus, the sum ofangles at each vertex is 240◦

(c) The sum of angles at each vertex is 270◦

(d) There are five equilateral triangles at each vertex so that the sum of theangles at each vertex is 300◦

(e) The faces consist of pentagons so the measure of each interior angle is108◦. The number of pentagons at each vertex is three so that the sum of theangles at each vertex is 324◦

Problem 41.14Determine for each of the following the minimum number of faces possible:(a) Prism (b) Pyramid (c) Hexahedron

Solution.(a) Five ( triangular base) (b) Four (triangular base) (c) Six

Problem 41.15True or false?(a) Through a given point not on plane P, there is exactly one line parallelto P.(b) Every set of four points is contained in one plane.(c) If a line is perpendicular to one of two parallel planes then it is perpen-dicular to the other.

Solution.(a) False. Any line through the point not crossing P is parallel to P(b) False. Three distinct points determine a unique plane. Take the fourthpoint oustide that plane.(c) True

Problem 41.16Tell whether each of the following suggests a polygon or a polygonal region.(a) A picture frame.(b) A page in this book.(c) A stop sign.

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Solution.(a) Polygon(b) Polygonal region(c) Polygonal region

Problem 41.17A certain prism has 20 vertices. How many faces and edges does it have?

Solution.Decagonal prism has 20 vertices, 12 faces (12 regular pentagons) and 30edges

Problem 41.18Why are there no such things as skew planes?

Solution.Two distinct planes must be parallel or intersecting

Problem 41.19A prism has a base with n sides.


Solution.(a) n + 2(b) 2n(c) 3n

Problem 41.20A pyramid has a base with n sides.


Solution.(a) n + 1 faces(b) n + 1 vertices(c) 2n

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Problem 42.1A small bottle of Perrier sparking water contains 33 cL. What is the volumein mL?

Solution.We have: 1 cL = 10 mL so that 33 cL = 330 mL

Problem 42.2Fill in the blanks.(a) 58728 g = kg(b) 632 mg = g(c) 0.23 kg = g

Solution.(a) 58728 g = 58728 g × 1 kg

1000 g= 58.728 kg

(b) 632 mg = 632 mg × 1 g1000 mg

= 0.632 g

(c) 0.23 kg = 0.23 kg × 1000 g1 kg

= 230 g

Problem 42.3Convert each of the following.(a) 100 in = yd(b) 400 yd = in(c) 300 ft = yd(d) 372 in = ft

Solution.(a) 100 in = 100 in× 1 ft

12 in× 1 yd

3 ft= 100

36= 25

9yd

(b) 400 yd = 400 yd× 3 ft1 yd

× 12 in1 ft

= 14400 in

(c) 300 ft = 300 ft× 1 yd3 ft

= 100 yd

(d) 372 in = 372 in× 1 ft12 in

= 31 ft


Item m cm mmLength of a piece of paper 35Height of a woman 1.63Width of a filmstrip 35Length of a cigarette 10

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Solution.

Item m cm mmLength of a piece of paper 0.35 35 350Height of a woman 1.63 163 1630Width of a filmstrip 0.035 3.5 35Length of a cigarette 0.1 10 100

Problem 42.5List the following in decreasing order: 8 cm, 5218 mm, 245 cm, 91 mm, 6 m,700 mm.

Solution.We have: 8 cm = 80 mm, 245 cm = 2450 mm, 6 m =6000 mm. Thus

8 cm < 91 mm < 700 mm < 2450 mm < 5218 mm < 6 m

Problem 42.6Complete each of the following.(a) 10 mm = cm(b) 3 km = m(c) 35 m = cm(d) 647 mm = cm

Solution.(a) 10 mm = 10 mm× 1 cm

10 mm= 1 cm

(b) 3 km = 3 km× 1000 m1 km

= 3000 m(c) 35 m = 35 m× 100 cm

1 m= 3500 cm

(d) 647 mm = 647 mm× 1 cm10 mm

= 64.7 cm

Problem 42.7Complete the following conversions.(a) 3 feet = inches(b) 2 miles = feet(c) 5 feet = yards

Solution.(a) 3 ft = 3 ft× 12 in

1 ft= 36 in

(b) 2 mi = 2 mi× 5280 ft1 mi

= 10560 ft

(c) 5 ft = 5 ft× 1 yd3 ft

= 53

ft

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Problem 42.8Complete the following conversions.(a) 7 yards = feet(b) 9 inches = feet(c) 500 yards = miles

Solution.(a) 7 yd = 7 yd× 3 ft

1 yd= 21 ft

(b) 9 in = 9 in× 1 ft12 in

= 0.75 ft

(c) 500 yd = 500 yd× 1 mi5280 ft

× 3 ft1 yd

= 2588

mi

Problem 42.9Complete the following conversions.(a) 9.4 L = mL(b) 37 mg = g(c) 346 mL = L

Solution.(a) 9.4 L = 9.4 L× 1000 mL

1 L= 9400 mL

(b) 37 mg = 37 mg × 1 g1000 mg

= 0.037 g

(c) 346 mL = 346 mL× 1 L1000 mL

= 0.346 L

Problem 42.10A nurse wants to give a patient 0.3 mg of a certain drug. The drug comes ina solution containing 0.5 mg per 2 mL. How many milliters should be used?

Solution.The nurse must use

0.3× 2

0.5= 1.2 mL

Problem 42.11A nurse wants to give a patient 3 gm of sulfisoxable. It comes in 500 mgtablets. How many tablets should be used?

Solution.Since 3 g = 3000 mg and 3000÷ 500 = 6 then the nurse must use 6 tablets

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Problem 42.12Complete the following conversions.(a) 3 gallons = quarts(b) 5 cups = pints(c) 7 pints = quarts(d) 12 cups = gallons

Solution.(a) 3 gallons = 3 gallons× 4 quarts

1 gallon= 12 quarts

(b) 5 cups = 5 cups× 1 pint2 cups

= 2.5 pints

(c) 7 pints = 7 pints× 1 quart2 pints

= 3.5 quarts

(d) 12 cups = 12 cups× 1 pint2 cups

× 1 quart2 pints

× 1 gallon4 quarts

= 0.75 gallons

Problem 42.13True or false? Expain.(a) 1 mm is longer than 1 in.(b) 1 m is longer than 1 km.(c) 1 g is heavier than 1 lb.(d) 1 gallon is more than 1 L.

Solution.(a) Since 1 in = 2.54 cm = 25.4 mm then the statement is false(b) Since 1 km = 1000 m then the statement is false(c) Since 1 lb = 16 oz = 16× 29 g = 464 g then the statement is false(d) Since 1 gal ≈ 3.79 L then the statement is true

Problem 42.14Derive a conversion formula for degrees Celsius to degrees Fahrenheit.

Solution.F is a linear function of C. That is, F = mC + b. But F = 32 when C = 0so that b = 32. Also, F = 212 when C = 100 so that m = 212−32

100= 9

5. Thus,

F =9

5C + 32

Problem 42.15A temperature of −10◦C is about(a) −20◦F (b) 10◦F (c) 40◦F (d) 70◦F

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Solution.F = −10× 9

5+ 32 = 14◦F so that (b) is the closest choice

Problem 42.16Convert the following to the nearest degree(a) Moderate oven (350◦F ) to degrees Celsius.(b) 20◦C to degrees Fahrenheit.(c) −5◦C to degrees Fahrenheit.

Solution.(a) 350 = 9

5C + 32. Solving for C we find C ≈ 177◦C

(b) F = 95× 20 + 32 = 68◦F

(c) F = 95× (−5) + 32 = 23◦F

Problem 42.17Complete the following conversions.(a) 1 cm2 = mm2

(b) 610 dam2 = hm2

(c) 564 m2 = km2

(d) 0.382 km2 = m2

Solution.(a) Since 1 cm = 10 mm then 1 cm2 = 100 mm2

(b) Since 1 dam = 0.1 hm then 1 dam2 = 0.01 hm2 and therefore 610 dam2

= 6.1 hm2

(c) Since 1 m = 0.001 km then 1 m2 = 0.000001 km2 so 564 m2 = 0.000564 km2

(d) Since 1 km = 1000 m then 1 km2 = 106 m2 so that 0.382 km2 = 382000m2

Problem 42.18Suppose that a bullet train is traveling 200 mph. How many feet per secondsis it traveling?

Solution.We have

200 mi

1 hr=

200 mi

1 hr× 5280 ft

1 mi× 1 hr

3600 sec≈ 203.3ft/sec

427

Problem 42.19A pole vaulter vaulted 19ft41

2in. Find the height in meters.

Solution.Converting to inches we find: 19ft41

2in = 19× 12 + 9

2= 232.5 in. But 1 in =

2.54 cm so that 232.5 in = 232.5× 2.54 = 590.55 cm = 5.9055 m

Problem 42.20The area of a rectangular lot is 25375 ft2. What is the area of the lot inacres? Use the fact that 640 acres = 1 square mile.

Solution.since 1 mi = 5280 ft then 1 mi2 = 27878400 ft2. Thus, 25375 ft2 =

2537527878400

mi2 = 2537527878400

× 640 ≈ 0.58 acres qed

Problem 42.21A vase holds 4286 grams of water. What is the capacity in liters? Recallthat the density of water is 1 g/cm3.

Solution.Since density = mass

volumethen the capacity of 4286 g in cm3 is 4286 cm3. But

1 cm3 = 1 mL = 0.001 L. Thus, the capacity is liters is 4286 × 0.001 =4.286 L

Problem 42.22By using dimensional analysis, make the following conversions.(a) 3.6 lb to oz(b) 55 mi/hr to ft/min(c) 35 mi/hr to in/sec(d) $575 per day to dollars per minute.

Solution.(a) 3.6 lb = 3.6 lb× 16 oz

1 lb= 57.7 oz

(b) 55 mi/hr = 55 mi1 hr

× 5280 ft1 mi

× 1 hr60 min

= 4840 ft/min

(c) 35 mi/hr = 35 mi1 hr

× 5280 ft1 mi

× 12 in1 ft

× 1 hr3600 sec

= 616 in/sec

(d) $ 575 per day = $5751 day

× 1 day1440 min

≈ 0.40 dollars per minute

Problem 42.23The density of a substance is the ratio of its mass to its volume. A chunk ofoak firewood weighs 2.85 kg and has a volume of 4100 cm3. Determine thedensity of oak in g/cm3, rounding to the nearest thousandth.

428

Solution.We have

2.85 kg

4100 cm3=

2.85 kg

4100 cm3× 1000 g

1 kg≈ 0.695 g/cm3

Problem 42.24The speed of sound is 1100 ft/sec at sea level. Express the speed of sound inmi/hr.

Solution.Using dimensional analysis we find

1100 ft/sec =1100 ft

1 sec× 1 mi

5280 ft× 3600 sec

1 hr= 750 mi/hr

Problem 42.25What temperature is numerically the same in degrees Celsius and degreesFahrenheit?

Solution.We are asked to solve the equation C = 9

5C + 32.

C = 95C + 32

−32 = 95C − C

45C = −32C = −40

Problem 43.1An oval track is made up by erecting semicircles on each end of a 50-m by100-m rectangle as shown in the figure below.

What is the perimeter of the track?

429

Solution.The perimter of the rectangular part is 2 × 100 + 2 × 50 = 300 m. Sincethe width of the rectangle is 50 m then the radius of each semi-circle is 25m. When the two half circles are joing they form a circle with perimeter2π(25) = 50π m. Hence, the perimeter of the oval track is p = 300 + 50π −100 = 200 + 50π m

Problem 43.2Find each of the following:(a) The circumference of a circle if the radius is 2 m.(b) The radius of a circle if the circumference is 15π m.

Solution.(a) C = 2πr = 4π m(b) r = C

2π= 15π

2π= 7.5 m

Problem 43.3Draw a triangle ABC. Measure the length of each side. For each of thefollowing, tell which is greater?(a) AB + BC or AC(b) BC + CA or AB(c) AB + CA or BC

Solution.(a) AB + BC > AC(b) BC + CA > AB(c) AB + CA > BC

Problem 43.4Can the following be the lengths of the sides of a triangle? Why or Why not?(a) 23 cm, 50 cm, 60 cm(b) 10 cm, 40 cm, 50 cm(c) 410 mm, 260 mm, 14 cm

Solution.According to the previous problem, the length of one side of a triangle is lessthan the sum of the lenghts of the remaining two sides.(a) The given numbers can be the lengths of the sides of a triangle.(b) Since 50 cm = 10 cm + 40 cm then the given numbers can not be the

430

lengths of the sides of a triangle.(c) Since 410 > 260 + 14 then the given numbers can not be the lengths ofthe sides of a triangle

Problem 43.5Find the circumference of a circle with diameter 6π cm.

Solution.We have C = 2πr = πd = π(6π) = 6π2 cm

Problem 43.6What happens to the circumference of a circle if the radius is doubled?

Solution.If the radius is doubles then the new circumference is 4πr. That is, the newperimeter is also doubled

Problem 43.7Find the length of the side of a square that has the same perimeter as arectangle that is 66 cm by 32 cm.

Solution. Let a be the length of the side of the square. Then 4a = 2(66 +32) = 4× 49. Divide both sides by 4 to obtain a = 49 cm

Problem 43.8A bicycle wheel has a diameter of 26 in. How far a rider travel in one fullrevolution of the tire? Use 3.14 for π.

Solution.We are asked to find the perimeter of the wheel. Since the wheel is a circlewith diameter 26 in then its circumference is C = πd = 26π = 26 × 3.14 =81.64 in

Problem 43.9A car has wheels with radii of 40 cm. How many revolutions per minutemust a wheel turn so that the car travels 50 km/h?

Solution.Using dimensional analysis we find

50 km

1 hr=

50 km

1 hr× 1 hr

60 min× 1 rev

0.0008π km≈ 332 rev/min

431

Problem 43.10A lot is 21 ft by 30 ft. To support a fence, an architect wants an uprightpost at each corner and an upright post every 3 ft in between. How many ofthese posts are needed?

Solution.Draw a picture. There are 22 posts to cover the lengths of the rectangle and12 posts to cover the widths a total of 34 posts

Problem 43.11Convert each of the following:(a) 1 cm2 = mm2

(b) 124,000,000 m2 = km2

Solution.(a) Since 1 cm = 10 mm then 1 cm2 = 100 mm2

(b) Since 1 m = 0.001 km then 124000000 m2 = 124000000 × 0.000001 =124 km2

Problem 43.12Find the cost of carpeting a 6.5 m × 4.5 m rectangular room if one metersquare of carpet costs $ 13.85.

Solution.The total area is 6.5 × 4.5 = 29.25 m2. Each square meter costs $ 13.85 sothe total cost is 29.25× 13.85 ≈ $405.11

Problem 43.13A rectangular plot of land is to be seeded with grass. If the plot is 22 m ×28 m and 1-kg bag of seed is needed for 85 m2 of land, how many bags ofseed are needed?

Solution.The area of the plot is 22× 28 = 616 m2 so that 616

85=≈ 7.2 then we need 8

bags of seed

432

Problem 43.14Find the area of the following octagon.

Solution.A hexagon consists of 8 equilateral triangle. According to the figure, eachside of the triangle is 4 cm and the height is 2

√3 cm. The area of a triangle is

4×2√

32

= 4√

3. Hence, the total area of the hexagon is 8×4√

3 = 32√

3 cm2

Problem 43.15(a) If a circle has a circumference of 8π cm, what is its area?(b) If a circle of radius r and a square with a side of length s have equalareas, express r in terms of s.

Solution.(a) The radius of the circle is r = C

2π= 8π

2π= 4 cm. Thus, its area is

A = πr2 = 16π cm2

(b) We have πr2 = s2. Solving for r we find r = s√

ππ

Problem 43.16A circular flower bed is 6 m in diameter and has a circular sidewalk aroundit 1 m wide. Find the area of the sidewalk.

Solution.The radius of the flower bed is 3 m. The radius of the circle consisting ofthe flower bed and the sidewalk is 4 m. Thus, the area of the sidewalk is16π − 9π = 7π m2

Problem 43.17(a) If the area of a square is 144 cm2, what is its perimeter?(b) If the perimeter of a square is 32 cm, what is its area?

Solution.(a) The side of the square is s =

√144 = 12 cm. The perimeter of the square

is p = 4s = 48 cm(b) The side of the square is s = 32

4= 8 cm. Its area is A = s2 = 64 cm2

433

Problem 43.18Find the area of each of the following shaded parts. Assume all arcs arecircular. The unit is cm.

Solution.(a) The area of the larger circle is 4π cm2. The area of each of the smallercircles is π cm2. Thus, the shaded area is 4π − 2π = 2π cm2

(b) The area of the half circle is 0.5π cm2 and the area of the triangle is2×22

= 2 cm2. Hence, the shaded area is 2 + 0.5π cm2

(c) The area is twice the area of the circle of radius 5 cm minus the area ofthe square of side 10 cm. That is, 50π − 100 cm2

Problem 43.19For the drawing below find the value of x.

Solution.By the Pythagorean formula we have: 72 = x2 + 52 or x2 + 25 = 49. Thus,x2 = 24 and x = 2

√6

Problem 43.20The size of a rectangular television screen is given as the length of the diag-onal of the screen. If the length of the screen is 24 cm and the width is 18cm, what is the diagonal length?

Solution.By the Pythagorean formula, the length of the diagonal is

√182 + 242 =

30 cm

434

Problem 43.21If the hypotenuse of a right triangle is 30 cm long and one leg is twice aslong as the other, how long are the legs of the triangle?

Solution.Let x be the length of one side. Then by the Pythagorean formula we have

x2 + (2x)2 = 302 = 900

Sovling for x we find 5x2 = 900 or x2 = 180. Hence, x = 6√

5 cm

Problem 43.22A 15-ft ladder is leaning against a wall. The base of the ladder is 3 ft fromthe wall. How high above the ground is the top of the ladder?

Solution.Let x be the distance from the top of the ladder to the ground. Then by thePythagorean formula we have

x2 + 32 = 152

Thus, x2 + 9 = 225 or x2 = 216. Hence, x =√

216 = 6√

6 ft

Problem 43.23Find the value of x in the following figure.

Solution.Let’s look at the triangle. The hypotenuse is 14 cm, one leg is x cm and theother leg is 25 - 15 = 10 cm. Using the Pythagorean formula we can writex2 + 102 = 142 or x2 + 100 = 196. Thus, x2 = 96 or x = 4

√6 cm

435

Problem 43.24Find x and y in the following figure.

Solution.Using the Pythagorean formula we can write x2 = 42 + (4

√3)2 = 64. Thus,

x = 8. On the other hand, y×82

= 4√

3×42

. Solving for y we find y = 2√

3

Problem 43.25Complete the following table which concerns circles:

Radius Diameter Circumference Area5 cm

24 cm17π m2

20πcm

Solution.

Radius Diameter Circumference Area5 cm 10 cm 10π cm 25π cm2

12 cm 24 cm 24π cm 144π cm2√

7 cm 2√

7 cm 2√

7π cm 17π cm2

10 cm 20 cm 20πcm 100π cm2

Problem 43.26Suppose the housing authority has valued the house shown here at $220 perft2. Find the assessed value.

436

Solution.The total area of the house is: 24 × 10 + 22 × 10 + 18 × 10 = 640 ft2. Theassessed value of the house is 640× 220 = $140800

Problem 43.27Two adjacent lots are for sale. Lot A cost $ 20,000 and lot B costs $27,000.Which lot has the lower cost per square meter?

Solution.The area of lot A is 42×30

2= 630 m2 whereas that of lot B is (24+30)×35

2=

945 m2. The cost of lot A per square meter is 20000630

≈ $31.75. The cost oflot B per square meter is 27000

945≈ $28.57. Thus, lot B has the lower cost per

square meter

Problem 43.28If the radius of a circle increases by 30% then by what percent does the areaof the circle increase?

Solution.If A = πr2 and r increases by 30% then the new area is π(1.3r)2 = 1.69πr2

so the area increased by 69%


Solution.The area of a quarter of the circle is 9π. The area of the triangle is 18. Thusthe shaded area is 9π − 18

437


Solutio.The diameter of the circle is: d2 = 52 + 122 = 25 + 144 = 169. Thus, d = 13.The area of half of the circle is 169

2π. The area of the triangle is 5×12

2= 30.

Thus the shaded area is 1692

π − 30

Problem 44.1A small can of frozen orange juice is about 9.5 cm tall and has a diameterof about 5.5 cm. The circular ends are metal and the rest of the can iscardboard. How much metal and how much cardboard are needed to makea juice can?

Solution.If we cut the lateral face and make it flat we will get a rectangle of height9.5 cm and length 5.5π. Thus, the area of the cardboard is (9.5) · (5.5π) =52.25π cm2.Tne metal ends each is a circle of radius 2.75 cm and thereforewith area π(2.75)2 cm2. So twice that is the area of the two metal, i.e.,15.125 cm2

Problem 44.2A pyramid has a square base 10 cm on a side. The edges that meet the apexhave length 13 cm. Find the slant height of the pyramid, and then calculatethe total surface area of the pyramid.

Solution.The base of the pyramid has area B = 100 cm2 and perimeter P = 40cm.The slant height can be computed with the Pythagorean formula whichgives l =

√132 − 52 = 12 cm. Thus, the surface area of the pyramid is

SA = 100 +1

2· 40 · 12 = 340 cm2

438

Problem 44.3An ice cream cone has a diameter of 2.5 in and a slant height of 6 in. Whatis the lateral surface area of the cone?

Solution.The lateral surface area is πrl where r is the radius of the base circle and lis the slant height. Thus, the lateral surface area is π ·

(2.52

)· 6 = 7.5π in2

Problem 44.4The diameter of Jupiter is about 11 times larger than the diameter of theplanet Earth. How many times greater is the surface area of Jupiter?

Solution.Let SJ be the surface area of Jupiter and SE that of the Earth. Then

SJ

SE

=4πr2

J

4πr2E

=

(dJ

2

)2(dE

2

)2 =d2

J

d2E

=121d2

E

d2E

= 121

Thus, the surface area of Jupiter is 121 times larger than the surface area ofthe Earth

Problem 44.5Find the surface area of each of the right prisms below.

Solution.(a) SA = 2 · 10 · 6 + 2 · 10 · 4 + 2 · 4 · 6 = 248 cm2

(b) Note that the base is a right triangle with hypotenuse of length 5 cm.The area of the base is then 3×4

2= 6 cm2. Thus,

SA = 2(6) + (3 + 4 + 5) · 10 = 132 cm2

439

Problem 44.6The Great Pyramid of Cheops is a right square pyramid with a height of 148m and a square base with perimeter of 930 m. The slant height is 188 m. Thebasic shape of the Transamerica Building in San Fransisco is a right squarepyramid that has a height of 260 m and a square base with a perimeter of140 m. The altitude of slant height is 261 m. How do the lateral surfaceareas of the two structures compare?

Solution.The length of one side of the square base of the Great Pyramid is 930

4=

232.5 m. The length of one side of the square base of the TransamericaBuilding is 35 m. The lateral surface area of each is then

LSA of Great Pyramid = 4 · 1

2· (232.5) · 188 = 87420 m2

and

LSA of Transamerica = 4 · 1

2· 35 · 261 = 18270 m2

Thus, the lateral surface area of the Great Pyramid is about 4.8 times largerthan that of the Transamerica Building

Problem 44.7Find the surface area of the following cone.

Solution.The base of the cone is a circle with radius 3 cm and area 9π cm2. Thelateral surface area is π · 3 · 5 = 15π cm2. Thus, the total surface area isSA = 9π + 15π = 24π cm2

Problem 44.8The Earth has a spherical shape of radius 6370 km. What is its surface area?

440

Solution.The surface area of the Earth is SA = 4πr = 4π(6370) = 25480π km2

Problem 44.9Suppose one right circular cylinder has radius 2 m and height 6 m and an-other has radius 6 m and height 2 m.

(a) Which cylinder has the greater lateral surface area?(b) Which cylinder has the greater total surface area?

Solution.(a) The lateral surface area of the first cylinder is 2π · 2 · 6 = 24π m2 andthat of the second cylinder is 2π ·6 ·2 = 24π m2. Thus, the two cylinder havethe same lateral surface area.(b) The surface area of the first cylinder is 2π · 2 · (2 + 6) = 32π m2 and thatof the second cylinder is 2π ·6 · (6+2) = 96π m2. Thus, the total surface areaof the first cylinder smaller than the total surface are of the second cylinder

Problem 44.10The base of a right pyramid is a regular hexagon with sides of length 12 m.The height of the pyramid is 9 m. Find the total surface area of the pyramid.

Soluion.The base is a hexagon with consists of 6 equilateral triangles each with sidesof length 12 m. The height in any such a triangle is found by the Pythagoreanformula to be 6

√3 m. Thus, the area of each triangle is 6

√3·122

= 36√

3 m2.

Hence, the area of base is 6 · 36√

3 = 216π m2. A face of the pyramid is arectangle of length 12 m and width of 9 m so that the area of a face is 108 m2.Hence, the total surface area of the pyramid is

SA = 6 · 108 + 2 · 216π = 648 + 432π m2

Problem 44.11A square piece of paper 10 cm on a side is rolled to form the lateral surfacearea of a right circular cylinder and then a top and bottom are added. Whatis the surface area of the cylinder?

441

Solution.The height of the cylinder is 10 cm. The radius of the base is found is follows:2πr = 10 or r = 5

π. Thus, the total surface area of the cylinder is

SA = 2π · 5

π·(

5

π+ 10

)=

50

π+ 100 cm2

Problem 44.12The top of a rectangular box has an area of 88 cm2. The sides have areas 32cm2 and 44 cm2. What are the dimensions of the box?

Solution.Let L be the length, W the width, and h the height. We have LW = 88, Lh =32, and Wh = 44. From the last two equations we find L = 8

11W. Thus, using

the first equation we find 811

W 2 = 88. Solving for W we find W = 11 cm.Thus, L = 8

11· 11 = 8 cm and h = 44

11= 4 cm

Problem 44.13What happens to the surface area of a sphere if the radius is tripled?

Solution.The surface area of a sphere is 4πr2. If the radius is tripled then the surfacearea becomes 4π(3r)2 = 36πr2. This shows that the surface area is 9 timeslarger

Problem 44.14Find the surface area of a square pyramid if the area of the base is 100 cm2

and the height is 20 cm.

Solution.Since the area of the base is 100 then the side of the square is 10 cm. Usingthe Pythagorean formula we find that the slant height is l =

√202 + 52 =√

425 = 5√

17. Thus, the surface area of the pyramid is

SA = 100 + 20 · 5√

17 = 100(1 + sqrt17) cm2

Problem 44.15Each region in the following figure revolves about the horizontal axis. Foreach case, sketch the three dimensional figure obtained and find its surface

442

area.

Solution.(a) The three dimensional shape is a cone with radius of base 10 cm andheight 20 cm. Its surface area is

SA = π · 10(10 +√

100 + 400) = 10π(10 + 10√

5) = 100π(1 +√

5) cm2

(b) The three dimensional shape is a cylinder of height 30 cm and radius ofbase 15 cm. Its surface area is

SA = 2π · 15(15 + 30) = 1350π cm2

Problem 44.16The total surface area of a cube is 10,648 cm2. Find the length of a diagonalthat is not a diagonal of a face.

Solution.The surface area of a cube of side s is 6s2. Thus, 6s2 = 10648 or s2 = 5324

2.

Hence, s =√

53242≈ 42 cm. The length of a diagonal that is not a diagonal

of a face is found by using the Pythagorean formula:

d =√

2 · 422 + 422 = 42√

3 ≈ 73 cm

Problem 44.17If the length, width, and height of a rectangular prism is tripled, how doesthe surface area change?

Solution.The surface area of a rectangular prism is given by the formula

SA = 2lw + 2h(l + w)

If we triple the length, width and height then the new SA is

SA = 2(3l)(3w) + 2(3h)(3l + 3w) = 9(2lw + 2h(l + w))

Thus the new surface area is 9 times larger then the previous one

443

Problem 44.18Find the surface area of the following figure.

Solution.The surface area of the cylinder is

SA = 2π · 3(3 + 10) = 78π cm2

The lateral surface are of the cone is

LSA = π · 3√

9 + 100 = 3π√

109 cm2

Thus, the total surface area is

78π + 3π√

109 cm2

Problem 44.19A room measures 4 meters by 7 meters and the ceiling is 3 meters high. Aliter of paint covers 40 square meters. How many liters of paint will it taketo paint all but the floor of the room?

Solution.The total surface area to be painted is

SA = 4× 7 + 2× 3× 4 + 2× 3× 7 = 94 m2

Since 9440

= 2.35 then 3 liters of paint should be purchased

Problem 44.20Given a sphere with diameter 10 , find the surface area of the smallest cylindercontaining the sphere.

444

Solution.The smallest cylinder containing the sphere has height 10 and radius of base5. Thus, the surface area of the cylinder is

SA = 2π · 5(5 + 10) = 150π


Solution.(a) V = 63 = 216 cm3

(b) V = 3 · 10 · 15 = 450 cm3

(c) V = hπr2 = 10 · π · 25 = 250π cm3


Solution.(a) V = 1

3· 42 · 5 = 80

3π cm3

(b) Using the Pythagorean formula we find h =√

100− 36 = 8. Thus, V =13π62 × 8 = 288

3π

445

Problem 45.3Maggie is planning to build a new one-story house with floor area of 2000ft2. She is thinking of putting in a 9-ft ceiling. If she does this, how manycubic feet of space will she have to heat or cool?

Solution.The house has the shape of rectangular prism with volume equals to 2000×9 = 18, 000 ft3

Problem 45.4Two cubes have sides lengths 4 cm and 6 cm, respectively. What is the ratioof their volumes?

Solution.The volume of the cube with side of length 4 cm is V4 = 4

3π · 43 = 256

3π. The

volume of the cube with side of length 6 cm is V6 = 43π · 63 = 864

3π. Thus,

V4

V6

=2563

π8643

π=

256

864=

8

27

Problem 45.5What happens to the volume of a sphere if its radius is doubled?

Solution.The volume of s a sphere of radius r is given by V = 4

3πr3. If we double the

radius then the volume of the new sphere is V ′ = 43π(2r)3 = 32

3πr3 = 8V.

That is the new volume is three times larger than the old volume

Problem 45.6An olympic-sized pool in the shape of a right rectangular prism is 50 m ×25 m. If it is 2 m deep throughout, how many liters of water does it hold?Recall that 1 m3 = 1000 L.

Solution.The volume of the prism is V = 50 × 25 × 2 = 2500 m3 = 2500 × 103 =2500000 L

Problem 45.7A standard straw is 25 cm long and 4 mm in diameter. How much liquid canbe held in the straw at one time?

446

Solution.The volume is given by V = 25π · 0.22 = π cm3 = π mL

Problem 45.8The pyramid of Khufu is 147 m high and its square base is 231 m on eachside. What is the volume of the pyramid?

Solution.The volume of the pyramid is V = 1

3· 2312 · 147 = 2614689 m3

Problem 45.9A square right regular pyramid is formed by cutting, folding, and gluing thefollowing pattern.

(a) What is the slant height of the pyramid?(b) What is the lateral surface area of the pyramid?(c) Use the Pythagorean formula to find the height of the pyramid.(d) What is the volume of the pyramid?

Solution.(a) Using the Pythagorean formula we find that the slant height is l =√

202 − 122 = 16 cm(b) LSA = 2 · 24 · 16 = 768 cm2

(c) By the Pythagorean formula the height is h =√

256− 144 =√

112 =4√

7 cm(d) V = 1

3· 242 × 4

√7 = 2304

3

√7 cm3

Problem 45.10A cube 10 cm on a side holds 1 liter. How many liters does a cube 20 cm ona side hold?

447

Solution.Since

V = 203 = 8000 cm3

then a cube 20 cm on a side hold 8 liters

Problem 45.11A right circular cone has height r and a circular base of radius 2r. Comparethe volume of the cone to that of a sphere of radius r.

Solution.The volume of the cone is VC = 1

3r · π · (2r)2 = 4

3πr3. The volume of the

sphere of radius r is VS = 43πr3. Thus, VC = VS

Problem 45.12A store sell two types of freezers. Freezer A costs $350 and measures 2 ft by2 ft by 4.5 ft. Freezer B cots $480 and measures 3 ft by 3 ft by 3.5 ft. Whichfreezer is the better buy?

Solution.The volume of freezer A is VA = 2 · 2 · 4.5 = 18 ft3 and that of freezer B isVB = 3 ·3 ·3.5 = 31.5 ft3. The cost per cubic foot of freezer A is 350

18≈ $19.44.

The cost per cubic foot of freezer B is 48031.5

≈ $15.24. Thus, Freezer B is thebetter buy

Problem 45.13Write a sentence that tells the difference between the surface area and volumeof a prism.

Solution.The surface area measures the total area of all its surfaces or faces, while thevolume measures the total amount of space in its interior

Problem 45.14A cylindrical water tank has a radius of 6.0 m. About how high must befilled to hold 400.0 m3?

Solution.The volume of the cylinder is V = h ·π · r2. Thus, h ·π ·36.0 = 400.0. Solvingfor h we find h = 400.0

36.0π≈ 3.5 m

448

Problem 45.15Roll an 8.5 by 11 in sheet of paper into a cylindrical tube. What is thediameter?

Solution.The cylinder is of height 11 and circumference of base equals to 8.5. Thus,πd = 8.5 and d = 8.5

π≈ 2.7

Problem 45.16A cylindrical pipe has an inner radius r, an outer radius R, and length l.Find its volume.

Solution.The volume of the pipe is πR2l − πr2l

Problem 45.17A basketball has a diameter of 10 in. What is its volume?

Solution.The volume is V = 4

3πr3 = 4

3π53 = 500

3π in3

Problem 45.18A standard tennis can is a cylinder that holds three tennis ball.

(a) Which is greater the circumference of the can or its height?(b) Find the volume of the can?

Solution.(a) Suppose that the tennis balls touch the sides, the top, and the bottom ofthe can. Let r be the radius of a tennis ball. Then the circumference of thecan is 2πr and the height is 6r. Since 2π > 6 then 2πr > 6r and thereforethe circumference of the can is larger than its height.(b) The volume of the can is V = (6r)(πr2) = 6πr3

Problem 45.19A cylindrical aquarium has a circular base with diameter 2 ft and height 3ft. How much water does it hold, in cubic feet?

Solution.The volume of the aquarium is V = 3 · π · 4 = 12π ft3

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Problem 45.20The circumference of a beach ball is 73 inches. How many cubic inches of airdoes the ball hold? Round your answer to the nearest cubic inch.

Solution.The radius of the ball is r = 73

2π. The volume of the ball is V = 4

3π

(732π

)3 ≈6569 in3

Problem 46.1Suppose ∆JKL ∼= ∆ABC, where ∆ABC is shown below.

Find the following(a) KL (b) LJ (c) m(∠L) (d) m(∠J)

Solution.(a) KL = BC = 4(b) LJ = CA = 5(c) m(∠L) = m(∠C) = 83◦

(d) m(∠J) = m(∠A) = 41◦

Problem 46.2Using congruence of triangles show that equilateral triangles are equiangular.

Solution.Let ABC be an equilateral triangles. Then ∆ABC ∼= ∆BAC by SSS prop-erty. This shows that m(∠A) = m(∠B). Similarly, ∆ABC ∼= ∆ACB bySSS. Hence, m(∠B) = m(∠C)

Problem 46.3Let the diagonals of a parallelogram ABCD intersect at a point M.(a) Show that ∆ABM ∼= ∆CDM.(b) Use part (a) to explain why M is the midpoint of both diagonals of theparallelogram.

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Solution.(a) We have AB = CD,m(∠ABM) = m(∠CDM), m(∠BAM) = m(∠DCM)(alternate interior angles). Hence, by ASA, ∆ABM ∼= ∆CDM.(b) By part (a), MA = MC so that M is the midpoint of AC. Similarly,MB = MD so that M is the midpoint of BD

Problem 46.4In the figure below, AB = AE and AC=AD.

(a) Show that m(∠B) = m(∠E)(b) Show that m(∠ACD) = m(∠ADC)(c) Show that ∆ABC ∼= ∆AED(d) Show that BC = DE.

Solution.(a) Since AB = AE then ∆ABE is isosceles so that m(∠B) = m(∠E)(b) Since AC = AD then ∆ACD is isosceles so that m(∠ACD) = m(∠ADC)(c) By (a) and (b) we have m(∠ACB) = m(∠ADE) and m(∠BAC) =m(∠DAE). Hence, by SAS property we have ∆ABC ∼= ∆AED(d) By (c) we have BC = DE

Problem 46.5Consider the following figure.

(a) Find AC(b) Find m(∠H), m(∠A), and m(∠C).

Solution.(a) By SAS property we have ∆ABC ∼= ∆FGH. Hence, AC = FH = 2.9 cm

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(b) m(∠H) = 180◦ − (62◦ + 78◦) = 40◦, m(∠A) = m(∠F ) = 78◦, m(∠C) =m(∠H) = 40◦

Problem 46.6Show that if ∆ABC ∼= ∆A′B′C ′ and ∆A′B′C ′ ∼= ∆A′′B′′C ′′ then ∆ABC ∼=∆A′′B′′C ′′.

Solution.Since ∆ABC ∼= ∆A′B′C ′ then AB = A′B′, AC = A′C ′, and BC = B′C ′.Similarly, since ∆A′B′C ′ ∼= ∆A”B”C” then A′B′ = A”B”, A′C ′ = A”C”,and B′C ′ = B”C”. Thus, AB = A”B”, AC = A”C”, and BC = B”C” sothat by SSS property ∆ABC ∼= ∆A”B”C”

Problem 46.7In the figure below, given that AB=BC=BD. Find m(∠ADC).

Solution.Triangle CBD is a right isosceles triangle with m(∠CBD) = 90◦. Hence,m(∠CDB) = 45circ. Similar argument shows that m(∠BDA) = 45◦. There-fore, m(∠ADC) = m(∠CDB) + m(∠BDA) = 90◦

Problem 46.8In the figure below given that AB = AC. Find m(∠A).

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Solution.Various angles are shown in the figure below.

Since AB = AC then triangle ABC is isosceles so that m(angleACB) =m(∠ABC). Hence, 180◦ − 8x + 3x = 4x. Solving this equation for x we findx = 18◦

Problem 46.9Find all missing angle measures in each figure.

Solution.

Problem 46.10An eighth grader says that AB=AC=AD , as shown in the figure below, thenm(∠B) = m(∠C) = m(∠D). Is this right? If not, what would you tell the

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child?

Solution.The conclusion is false since m(∠B) = m(∠BCA) < m(∠C)

Problem 46.11What type of figure is formed by joining the midpoints of a rectangle?

Solution.From the figure below we see that BB′ = CD′, CC ′ = BC ′ and m(∠B) =m(∠C) = 90◦. Thus, by the SAS property we conclude that ∆B′BC ′ ∼=D′CC ′ so that B′C ′ = C ′D′. Similar arguments show that B′C ′ = C ′D′ =A′B′ = A′D′ so that A′B′C ′D′ is a rhombus

Problem 46.12If two triangles are congruent what can be said about their perimeters? ar-eas?

Solution.Suppose that ∆ABC ∼= ∆A′B′C ′. Then AB = A′B′, AC = A′C ′, BC =B′C ′. Thus the two triangles have equal perimeters

Problem 46.13In a pair of right triangles, suppose two legs of one are congruent respectivelyto two legs of the other. Explain whether the triangles are congruent andwhy.

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Solution.The two triangles are congruent by the SAS property

Problem 46.14A rural homeowner had his television antenna held in place by three guywires, as shown in the following figure. If the distance to each of the stakesfrom the base of the antenna are the same, what is true about the lengths ofthe wires? Why?

Solution.The lengths must be the same because they are corresponding sides of con-gruent triangles

Problem 46.15For each of the following, determine whether the given conditions are suffi-cient to prove that ∆PQR ∼= ∆MNO. Justify your answer.(a) PQ = MN, PR = MO, m(∠P ) = m(∠M)(b) PQ = MN, PR = MO, QR = NO(c) PQ = MN, PR = MO, m(∠Q) = m(∠N)

Solution.(a) SAS property guarantees that the two triangles are congruent(b) SSS property guarantees that the two triangles are congruent(c) Conditions do not necessarily imply that the two triangles are congruent

Problem 46.16Given that ∆RST ∼= ∆JLK, complete the following statements.(a) ∆TRS ∼= ∆

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(b) ∆SRT ∼= ∆(c) ∆TSR ∼= ∆(d) ∆JKL ∼= ∆

Solution.(a) ∆TRS ∼= ∆KJL(b) ∆SRT ∼= ∆LJK(c) ∆TSR ∼= ∆KLJ(d) ∆JKL ∼= ∆RTS

Problem 46.17You are given ∆RST and ∆XY Z with m(∠S) = m(∠Y ).(a) To show ∆RST ∼= ∆XY Z by the SAS congruence property, what morewould you need to know?(b) To show that ∆RST ∼= ∆XY Z by the ASA congruence property, whatmore would you need to know?

Solution.(a) RS = XY and ST = Y Z(b) RS = XY and m(∠R) = m(∠X

Problem 46.18You are given ∆ABC and ∆GHI with AB = GH. To show that ∆ABC ∼=∆GHI by the SSS congruence property, what more would you need to know?

Solution.AC = GI and BC = HI

Problem 46.19Suppose that ABCD is a kite with AB = AD and BC = DC. Show that thediagonal AC divides the kite into two congruent triangles.

Solution.Since AB = AD, BC = DC, and AC = AC then ∆ABC ∼= ∆ADC by SSS

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Problem 46.20(a) Show that the diagonal of a parallelogram divides it into two congruenttriangles.(b) Use part (a) to show that the opposite sides of a parallelogram are con-gruent.(c) Use part (a) to show that the opposite angles of a parallelogram arecongruent.

Solution.(a) Since m(∠CAD) = m(∠BCD and m(∠ACD) = m(∠BAC then by ASA∆ABC ∼= ∆CDA. A similar argument shows that ∆BCD ∼= ∆BAD(b) By (a) AB = CD and BC = DA(c) By (a) m(∠B) = m(∠D) and m(∠A) = m(∠C

Problem 47.1Which of the following triangles are always similar?(a) right triangles(b) isosceles triangles(c) equilateral triangles

Solution.(a) No. For example, consider the two right triangles ∆ABC and ∆DEFwith m(∠A) = m(∠C) = 45◦, m(∠B) = 90◦, m(∠D) = 35◦, m(∠F = 55,and m(∠E) = 90◦

(b) Yes by the SAS property.(c) Yes by the AA property since all angles are the same

Problem 47.2Show that if ∆ABC ∼ ∆A′B′C ′ and ∆A′B′C ′ ∼ ∆A′′B′′C ′′ then ∆ABC ∼∆A′′B′′C ′′.

Solution.Suppose that ∆ABC ∼ ∆A′B′C ′ then m(∠A) = m(∠A′), m(∠B) = m(∠B′), m(∠C) =m(∠C ′) and

AB

A′B′ =AC

A′C ′ =BC

B′C ′

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Similarly, if ∆A′B′C ′ ∼ ∆A′′B′′C ′′ then m(∠A′) = m(∠A”), m(∠B′) =m(∠B”), m(∠C ′) = m(∠C”) and

A′B′

A”B”=

A′C ′

A”C”=

B′C ′

B”C”

Thus, m(∠A) = m(∠A”), m(∠B) = m(∠B”), m(∠C) = m(∠C”) and

AB

A”B”=

AC

A”C”=

BC

B”C”

This shows that ∆ABC ∼ ∆A′′B′′C ′′

Problem 47.3Each pair of triangles is similar. By which test can they be proved to besimilar ?

Solution.(a) SAS(b) AA(c) SSS

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Problem 47.4Suppose ∆ABC ∼ ∆DEF with scaled factor k.(a) Compare the perimeters of the two triangles.(b) Compare the areas of the two triangles.

Solution.(a) The perimeter of ∆ABC is AB + BC + CA. The perimeter of ∆DEF isDE + EF + FE. But DE = kAB, DF = kAC, and EF = kBC. Thus, theperimeter of ∆DEF is k times larger than the perimeter of ∆ABC.(b) Let h be a height of ∆ABC with base BC and h′ be a height if DEFwith base EF. Then

h·BC2

h′·EF2

= k2

Problem 47.5Areas of two similar triangles are 144 sq.cm. and 81 sq.cm. If one side of thefirst triangle is 6 cm then find the corresponding side of the second triangle.

Solution.

By the previous exercise the scaled factor is k =√

14481

= 43. Now, if we assume

∆ABC ∼ ∆DEF and AB = 6 cm then DE = kAB = 43· 6 = 12 cm

Problem 47.6The side of an equilateral triangle ∆ABC is 5 cm. Find the length of theside of another equilateral triangle ∆PQR whose area is four times area of∆ABC.

Solution.Since the two triangles are equilateral then ∆ABC ∼ ∆PQR. Since the areaof ∆PQR is four times the area of ∆ABC then k = 2. Since AB = 2PQthen PQ = 5

2= 2.5 cm

Problem 47.7The corresponding sides of two similar triangles are 4 cm and 6 cm. Findthe ratio of the areas of the triangles.

Solution.We are given that k = 6

4= 1.5. Thus, the ratio of the areas of the two

triangles is k2 = 94

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Problem 47.8A clever outdoorsman whose eye-level is 2 meters above the ground, wishesto find the height of a tree. He places a mirror horizontally on the ground20 meters from the tree, and finds that if he stands at a point C which is4 meters from the mirror B, he can see the reflection of the top of the tree.How high is the tree?

Solution.By the AA test we have ∆BPC ∼ ∆BQA. Thus,

QA

PC=

BA

BC=

20

4= 5.

Hence, QA = 5PC = 10 m

Problem 47.9A child 1.2 meters tall is standing 11 meters away from a tall building. Aspotlight on the ground is located 20 meters away from the building andshines on the wall. How tall is the child’s shadow on the building?

Solution.By the AA property ∆ABC ∼ ∆APQ. Thus,

AC

AQ=

BC

PQ

Hence,9

20=

1.2

PQ

Solving for Q we find PQ = 249≈ 2.67 m

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Problem 47.10On a sunny day, Michelle and Nancy noticed that their shadows were dif-ferent lengths. Nancy measured Michelle’s shadow and found that it was 96inches long. Michelle then measured Nancy’s shadow and found that it was102 inches long.

(a) Who do you think is taller, Nancy or Michelle? Why?(b) If Michelle is 5 feet 4 inches tall, how tall is Nancy?(c) If Nancy is 5 feet 4 inches tall, how tall is Michelle?

Solution.(a) Let M be the height of Michelle and N that of Nancy. Then using similartriangles, we have

M

N=

96

102< 1.

This shows that Nancy is taller than Michelle.(b) If M = 64 in then N = 64·102

96= 68 in. That is, Nancy is 5 feet 8 inches

tall.(c) If N = 64 in then M = 96·64

102≈ 60 in or about 5 feet

Problem 47.11An engineering firm wants to build a bridge across the river shown below.An engineer measures the following distances: BC = 1,200 feet, CD = 40feet, and DE = 20 feet.

(a) Prove ∆ABC is similar to ∆EDC.(b) Railings cost $4 per foot. How much will it cost to put railings on bothsides of the bridge?

Solution.(a) Since m(∠B) = m(∠D) = 90◦ and m(∠BCA) = m(∠DCE) (vertical

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angles) then by the AA test ∆ABC ∼ ∆EDC.(b) Finding the length of AB we have

AB

DE=

BC

CD

orAB

20=

1200

40

Solving for AB we find AB = 600 ft. Thus, it costs 1200 × 4$4, 800 to putrailings on both sides of the bridge

Problem 47.12At a certain time of the day, the shadow of a 5′ boy is 8′ long. The shadowof a tree at this same time is 28′ long. How tall is the tree?

Solution.Using similar triangles we can write

height of tree

height of boy=

shadow of tree

shadow of boy

That is,height of tree

5=

28

8

Hence, the height of the tree is 5 · 288

= 17.5′

Problem 47.13Two ladders are leaned against a wall such that they make the same anglewith the ground. The 10′ ladder reaches 8′ up the wall. How much furtherup the wall does the 18′ ladder reach?

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Solution.By the AA test the two triangles are similar so that

top of 18′ ladder from ground

8=

18

10

Hence, the top of 18’ ladder from the ground is 8 · 1810

= 14.4′

Problem 47.14Given that lines DE and AB are parallel in the figure below, determine thevalue of x, i.e. the distance between points A and D.

Solution.By the AA test we have ∆CED ∼ ∆CBA so that

CD

CA=

ED

AB

or15

x + 15=

7

11

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Solving this equation for x we find

15x+15

= 711

7(x + 15) = 15 · 117x + 105 = 165

7x + 105− 105 = 165− 1057x = 60x = 60

7≈ 8.57

Problem 47.15In the figure below, lines AC and DE are vertical, and line CD is horizontal.Show that ∆ABC ∼ ∆EBD.

Solution.We have: m(∠C) = m(∠D) = 90◦ and m(∠CBA) = m(∠DBE) (verticalangles). Hence, by the AA test we have ∆ABC ∼ ∆EBD

Problem 47.16Find a pair of similar triangles in each of these figures:

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Solution.∆ABC ∼ ∆ADE by the AA test. Similarly, ∆TUV ∼ ∆TSR

Problem 47.17Find x :

Solution.Since m(∠E) = m(∠D) = 90◦ and m(∠DBA) = m(∠EBC) (vertical angles)then ∆BDA ∼ ∆BEC. This implies that

DB

x=

10

6

or14− x

x=

10

6

Solving for x we find14−x

x= 10

6

10x = 6(14− x)10x = 84− 6x16x = 84x = 84

16= 5.25

Problem 47.18In the diagram, DE is parallel to AC. Also, BD = 4, DA = 6 and EC = 8.Find BC to the nearest tenth.

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Solution.By the AA test we have ∆BDA ∼ BAC so that

BC

BE=

BA

BD

orBC

BC − 8=

10

4

Solving this equation for BC we find

BCBC−8

= 104

4BC = 10(BC − 8)4BC = 10BC − 806BC = 80BC = 80

6= 13.3

Problem 47.19Find BC.

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Solution.See the previous exercise

Problem 47.20Find BE

Solution.Since ∆BDE ∼ ∆BAC then

BE

BC=

4

12

orBE

BE + 9=

4

12

Solving for BE we find

BEBE+9

= 13

3BE = BE + 92BE = 9BE = 4.5

Problem 47.21Copy and complete the given table. It is given that OH

HJ= OI

IK.

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Solution.We have: OJ = OH + HJ = 15 + 5 = 20. Also, 15

5= 45

IK. Solving for IK we

find IK = 453

= 15. Finally, OK = OI + IK = 45 + 15 = 60

OH HJ OJ OI IK OK15 5 20 45 15 15

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Documents

A Second Course in Mathematics Concepts for Elementary Teachers